Chemistry Question Paper
SECTION – A
Q.01 Alkali and alkaline earth metals along with hydrogen and helium constitute s-block elements. They have low ionization enthalpies and hence exhibit characteristic flame colourations.
(a) Which of the following elements do not impart any colour to the flame?
(1) Calcium (2) Sodium (3) Magnesium (4) Lithium
Answer: (3) Magnesium
Explanation: Magnesium does not typically impart a distinct color to a flame in a standard flame test. While it does burn, the emission is primarily in the UV range, not visible.
(b) The correct order of atomic size of the following elements is: Ca, Be, Mg
(1) Be > Mg > Ca (2) Ca > Mg > Be (3) Mg > Ca > Be (4) Ca > Be > Mg
Answer: (2) Ca > Mg > Be
Explanation: Atomic size increases down a group and decreases across a period (from left to right). Calcium is below Magnesium in the same group, and both are to the right of Beryllium in the same period.
(c) The anomalous behaviour of lithium is due to
(1) large size of lithium (2) small size of lithium (3) low ionization enthalpy of lithium (4) presence of d-orbital in lithium
Answer: (2) small size of lithium
Explanation: Lithium’s small size leads to its higher polarizing power, greater charge density, and stronger interatomic interactions compared to other alkali metals. This results in differences in properties like hydration enthalpy, electro negativity, and a tendency to form covalent compounds.
(d) Alkali metals are not found in nature because
(1) they are highly reactive (2) they are chemically inert (3) they have the largest sizes in a period (4) they have low ionization enthalpy.
Answer: (1) they are highly reactive
Explanation: Alkali metals readily lose their one valence electron to form +1 ions due to their low ionization energies. This makes them extremely reactive with other elements, so they are always found in compounds, never in their elemental (free) state.
Q.02 Chemical equilibria are important in numerous biological and environmental processes. The rates of forward and reverse reactions become equal at equilibrium. The magnitude of the equilibrium constant gives an idea of the relative amounts of reactants and products. Change in concentration, temperature or pressure will shift the equilibrium in a direction that tends to undo the effect of change imposed.
(i) The equilibrium constant value Kc changes for the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g) with
(a) Pressure
(b) Catalyst
(c) addition of PCl5
(d) temperature
Answer: (d) temperature
Explanation: Equilibrium constants are temperature-dependent. Changing pressure can shift the position of equilibrium (Le Chatelier’s principle), but it doesn’t change the value of the equilibrium constant (Kc or Kp) unless the temperature is also changed. Catalysts affect the rate of reaching equilibrium but do not change the equilibrium constant. Changing concentrations also shifts the position of equilibrium but not the equilibrium constant.
(ii) If the equilibrium constant for N2(g) + O2(g) ⇌ 2NO is K, the equilibrium constant for ½ N2(g) + ½ O2(g) ⇌ NO(g) will be
(a) K
(b) K½
(c) K2
(d) K1/2
Answer: (b) K½
Explanation: If the coefficients of a balanced chemical equation are multiplied or divided by a factor, the equilibrium constant is raised to the power of that factor. Here, the original equation is divided by 2, so the new equilibrium constant is K^(1/2) or √K.
OR
For the reaction CO(g) + Cl2(g) ⇌ COCl2(g), Kp/Kc is equal to:
(a) √RT
(b) RT
(c) 1.0
(d) 1/RT
Answer: (d) 1/RT
Explanation: Kp = Kc(RT)^(Δn), where Δn is the change in the number of moles of gas. Here, Δn = 1 – (1+1) = -1. So, Kp/Kc = (RT)^(-1) = 1/RT.
(iii) For the reaction, N2(g) + 3H2(g) ⇌ 2NH3(s) + heat, the equilibrium shifts in the forward direction
(a) by increasing the concentration of NH3
(b) by decreasing the pressure
(c) by decreasing the temperature
(d) By increasing the temperature
Answer: (c) by decreasing the temperature
Explanation: The reaction is exothermic (releases heat). Decreasing the temperature favors the exothermic reaction (Le Chatelier’s principle), shifting the equilibrium to the right (forward direction).
(iv) If the value of the equilibrium constant for a particular reaction is 1.6 x 10^12, then at equilibrium, the system will contain
(a) Mostly products
(b) mostly reactants
(c) all reactants
(d) similar amounts of reactants and products
Answer: (a) Mostly products
Explanation: A large equilibrium constant indicates that the equilibrium position lies far to the right, meaning that at equilibrium, the concentration of products is much greater than the concentration of reactants.
Q.03 For the process to occur under adiabatic conditions, the correct condition is:
(a) ΔT = 0
(b) ΔP = 0
(c) q = 0
(d) w = 0
Answer: (c) q = 0
Explanation: Adiabatic processes are defined by no heat exchange (q) between the system and surroundings.
Q.04 Which among the following will behave like a Lewis acid?
(a) BF₃
(b) H₂O
(c) OH⁻
(d) NH₃
Answer: (a) BF₃
Explanation: BF₃ is electron deficient, making it a Lewis acid (electron pair acceptor).
Q.05 Addition of water to ethyne in the presence of H₂SO₄ and mercuric sulphate gives:
(a) Ethanol
(b) Ethanal
(c) Ethanoic acid
(d) Ethane
Answer: (b) Ethanal
Explanation: The reaction follows Markovnikov’s rule, and the initial product is a vinyl alcohol which tautomerizes to ethanal.
OR
Tertiary butyl bromide on heating with alcoholic potash gives:
(a) But-1-ene
(b) 2-methylpropene
(c) Propene
(d) But-2-ene
Answer: (b) 2-methylpropene
Explanation: The reaction is an E2 elimination, and the major product is the more substituted alkene (Zaitsev’s rule).
Q.06 The enthalpies of all elements in their standard states are:
(a) unity
(b) less than zero
(c) zero
(d) different for each element
Answer: (c) zero
Explanation: By definition, the standard enthalpy of formation of an element in its standard state is zero.
Q.07 Which among the following is an electron precise hydride?
(a) water
(b) Ammonia
(c) Methane
(d) Borane
Answer: (c) Methane
Explanation: Methane (CH₄) is an electron-precise hydride, meaning it has the exact number of electrons needed for bonding (no excess or deficiency).
Q.08 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. The change in internal energy
(a) -307 J
(b) +307 J
(c) -1095 J
(d) +1095 J
Answer: (b) +307 J
Explanation: ΔU = Q – W = 701 J – 394 J = 307 J
OR
A reaction, A + B + C + D ⇌ E + F + q is found to have a positive entropy change. The reaction will be:
(a) possible at high temperature
(b) possible at any temperature
(d) not possible at any temperature
Answer: (a) possible at high temperature
Explanation: A positive entropy change favors spontaneity. Since ΔG = ΔH – TΔS, a positive ΔS makes ΔG more negative (favorable) with increasing temperature.
Q.09 For the process: 2Cl(g) → Cl₂(g)
(a) ΔH > 0, ΔS > 0
(b) ΔH > 0, ΔS < 0
(c) ΔH < 0, ΔS < 0
(d) ΔH < 0, ΔS > 0
Answer: (c) ΔH < 0, ΔS < 0
Explanation: Bond formation releases energy (ΔH < 0), and going from 2 particles to 1 decreases disorder (ΔS < 0).
Q.10 The alkene on ozonolysis gives acetone and propanone is:
(a) 2-methylbut-2-ene
(b) 3-methylbut-1-ene
(c) Pent-2-ene
(d) 2-methylpropene
Answer: (a) 2-methylbut-2-ene
Explanation: Ozonolysis cleaves the double bond. 2-methylbut-2-ene yields acetone and propanone upon ozonolysis.
Q.11 For a reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔrH°H = -92.4 kJ/mol, then the standard enthalpy of formation of NH₃(g) is:
(a) +46.2 kJ/mol
(b) -46.2 kJ/mol
(c) Zero
(d) -92.4 kJ/mol
Answer: (b) -46.2 kJ/mol
Explanation: The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states. Since the given reaction forms 2 moles of NH₃, the enthalpy of formation of 1 mole of NH₃ is (-92.4 kJ/mol) / 2 = -46.2 kJ/mol.
In the following questions (12-16), a statement of Assertion followed by a statement of Reason is given, choose the correct answer out of the
(A) Both Assertion and Reason are true and Reason is the correct eplanation of Assertion.
(B) Both Assertion and Reason
(C) Assertion is True but Reason is false.
(D) Both Assertion and Reason
Q.12 Assertion: Enthalpy is a state function.
Reason: The value of Enthalpy is path dependent.
Answer: (C) Assertion is True but Reason is false.
Explanation: Enthalpy is a state function, meaning its value depends only on the initial and final states, not the path taken.
Q.13 Assertion: If Qc (reaction quotient) is less than Kc (equilibrium constant), the reaction moves in the direction of reactants.
Reason: The value of the reaction quotient is independent of temperature.
Answer: (D) Both Assertion and Reason are false.
Explanation: If Qc < Kc, the reaction moves in the direction of products, not reactants, to reach equilibrium. The reaction quotient is dependent on temperature.
Q.14 Assertion: Ethyne is more acidic than ethene.
Reason: In ethyne, carbon is sp hybridized.
Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Explanation: sp hybridization in ethyne leads to greater s-character in the C-H bond, making the hydrogen more acidic.
Q.15 Assertion: Benzene does not decolorize bromine water.
Reason: Benzene is stabilized by resonance due to delocalization of π-electrons.
Answer: (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
Explanation: Benzene’s delocalized π-electrons make it less reactive towards electrophilic addition reactions like the addition of bromine water.
Q.16 Assertion: The boiling point of isopentane is higher than normal pentane.
Reason: Branching increases the boiling point of isomeric alkanes.
Answer: (D) Both Assertion and Reason are false.
Explanation: Branching decreases the boiling point of isomeric alkanes due to weaker van der Waals forces. Isopentane has a lower boiling point than n-pentane.
SECTION – B
Q.17 What characteristics do you expect from an e⁻ deficient hydride with respect to its structure?
Electron-deficient hydrides typically exhibit the following structural characteristics:
- Bridging: They often form bridge bonds (3-center-2-electron bonds) where hydrogen atoms share electrons with more than two atoms.
- Polymeric Structures: They tend to form polymeric structures to achieve electronic stability.
- Lewis Acidity: They act as Lewis acids, accepting electron pairs from Lewis bases.
- Incomplete Octet: The central atom in the hydride does not have a complete octet of electrons.
Q.18 Lithium shows similarities with magnesium in its chemical behavior. What is the cause of these similarities?
The similarity between Lithium and Magnesium is due to the diagonal relationship. This arises from the similar charge/size ratio (ionic potential) of Li⁺ and Mg²⁺ ions. This leads to similarities in polarizing power and thus in the properties of the compounds they form.
Q.19 Atomic radius of Gallium is less than that of aluminum. Why?
The smaller atomic radius of Gallium compared to Aluminum is due to the poor shielding effect of the 3d electrons in Gallium. The increased nuclear charge is not effectively shielded, leading to a stronger attraction for the valence electrons and a smaller atomic radius.
Q.20 Give the IUPAC name of the following compounds:
(a) (Structure with a carbonyl group (C=O) attached to a benzene ring and a hydroxyl group (OH))
The IUPAC name is Benzoic Acid.
(b) (Structure with a benzene ring attached to a carbon with a cyano group (CN))
The IUPAC name is Benzonitrile.
OR
Give the IUPAC name of the following compounds:
(a) (Structure with a chain of carbons, a methyl group (CH3) on the second carbon, and a hydroxyl group (CH2OH) on the end carbon)
The IUPAC name is 2-Methylpropan-1-ol.
(b) (Structure with a chain of carbons, a double bond between the second and third carbons, an aldehyde group (CHO) on the end carbon, and a chlorine (Cl) on the first carbon)
The IUPAC name is 1-Chlorohex-2-enal.
Q.21 Write bond line formula of:
(a) Isopropyl alcohol
\
CH3-CH-OH
/
CH3
(b) Heptan-4-one
O
||
CH3CH2CH2C CH2CH2CH3
Q.22 Write the conjugate acid and conjugate base for the following:
(a) HSO₄⁻
Conjugate acid: H₂SO₄ Conjugate base: SO₄²⁻
(b) H₂O
Conjugate acid: H₃O⁺ Conjugate base: OH⁻
Q.23 For the reaction at 298 K, 2A + B → C, ΔH = 400 kJ/mol and ΔS = 0.2 kJ/K/mol. At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
For a reaction to be spontaneous, ΔG must be negative.
ΔG = ΔH – TΔS
0 = ΔH – TΔS (at equilibrium, ΔG = 0)
T = ΔH/ΔS = (400 kJ/mol) / (0.2 kJ/K/mol) = 2000 K
The reaction will be spontaneous when T > 2000 K.
Q.24 Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are -110, -393, 81 and 9.7 kJ/mol respectively. Find the value of ΔrH standard enthalpy for the reaction:
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
ΔrH = [ΔfH(N₂O) + 3ΔfH(CO₂)] – [ΔfH(N₂O₄) + 3ΔfH(CO)]
ΔrH = [(81) + 3(-393)] – [(9.7) + 3(-110)]
ΔrH = [81 – 1179] – [9.7 – 330]
ΔrH = -1098 – (-320.3)
ΔrH = -777.7 kJ/mol
Q.25 Convert:
(a) Benzene to p-nitrobromobenzene
- Nitration: React benzene with a mixture of concentrated nitric acid and sulfuric acid to get nitrobenzene.
- Bromination: React nitrobenzene with Br₂ and FeBr₃ to get p-nitrobromobenzene. The nitro group is an ortho/para director, and para is favored due to less steric hindrance.
(b) Ethyne to Benzene
- Cyclic Polymerization: Pass ethyne through a red hot copper tube at about 873 K. This causes the ethyne molecules to undergo cyclic polymerization, resulting in the formation of benzene.
SECTION – C
Q.26 A sample of HI(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the
2HI(g)
Initial pressure of HI = 0.2 atm Equilibrium pressure of HI = 0.04 atm
Change in pressure of HI = 0.2 – 0.04 = 0.16 atm
Since 2 moles of HI give 1 mole of H₂ and 1 mole of I₂, the partial pressures of H₂ and I₂ at equilibrium will each be half the change in pressure of HI.
Partial pressure of H₂ = 0.16 atm / 2 = 0.08 atm Partial pressure of I₂ = 0.16 atm / 2 = 0.08 atm
Kp = (P_H₂ * P_I₂) / (P_HI)² = (0.08 * 0.08) / (0.04)² = 4
OR
One mole of H₂O and one mole of CO are taken in a 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation.
Initial moles of H₂O = 1 Initial moles of CO = 1 Volume = 10 L
At equilibrium, 40% of water reacts, so moles of H₂O reacted = 0.4 moles.
Equilibrium moles: H₂O = 1 – 0.4 = 0.6 CO = 1 – 0.4 = 0.6 H₂ = 0.4 CO₂ = 0.4
Equilibrium concentrations: [H₂O] = 0.6/10 = 0.06 M [CO] = 0.6/10 = 0.06 M [H₂] = 0.4/10 = 0.04 M [CO₂] = 0.4/10 = 0.04 M
Kc = ([H₂][CO₂]) / ([H₂O][CO]) = (0.04 * 0.04) / (0.06 * 0.06) = 0.44
Q.27 What do you expect the nature of hydrides if formed by elements of atomic numbers 15, 19, 23 with hydrogen?
- Atomic number 15 (Phosphorus): Forms covalent hydrides like PH₃ (phosphine). These are typically gases or volatile liquids and are not very stable.
- Atomic number 19 (Potassium): Forms ionic hydrides like KH (potassium hydride). These are typically solids with a salt-like structure and are highly reactive with water.
- Atomic number 23 (Vanadium): Forms metallic or interstitial hydrides. These are non-stoichiometric compounds where hydrogen atoms occupy the interstitial spaces in the metal lattice. They often have metallic properties.
Q.28 The stability of lower oxidation state increases while that of higher oxidation state decreases down the group in the p-block. Give reason.
This is due to the inert pair effect. As we go down the p-block, the tendency of the s-electrons in the valence shell to participate in bonding decreases. This is because the s-electrons are more strongly held by the nucleus due to the poor shielding effect of the intervening d and f electrons. As a result, the lower oxidation state (involving only p-electrons) becomes more stable, while the higher oxidation state (involving both s and p electrons) becomes less stable.
Q.29 For the reaction:
2A(g) + B(g) → 2D(s)
ΔU° (Internal energy) = -10.5 kJ and ΔS° (entropy) = -44.1 JK⁻¹ at 25°C
Calculate ΔG° for the reaction and predict whether the reaction may occur spontaneously.
ΔG° = ΔH° – TΔS°
First, we need to convert ΔU° to ΔH°:
ΔH° = ΔU° + Δn_g RT
Δn_g = moles of gaseous products – moles of gaseous reactants = 0 – (2+1) = -3
R = 8.314 J mol⁻¹ K⁻¹ T = 25°C + 273.15 = 298.15 K
ΔH° = -10.5 kJ + (-3)(8.314 x 10⁻³ kJ mol⁻¹ K⁻¹)(298.15 K) ≈ -17.93 kJ
Now, calculate ΔG°:
ΔG° = ΔH° – TΔS° = -17.93 kJ – (298.15 K)(-44.1 x 10⁻³ kJ K⁻¹) ≈ -4.75 kJ
Since ΔG° is negative, the reaction is spontaneous at 25°C.
OR
The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are -890.3 kJ/mol, -393.5 kJ/mol and -285.8 kJ/mol respectively.
Combustion reactions:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH°comb = -890.3 kJ/mol C(s) + O₂(g) → CO₂(g) ΔH°comb = -393.5 kJ/mol H₂(g) + ½O₂(g) → H₂O(l) ΔH°comb = -285.8 kJ/mol
Formation reaction:
C(s) + 2H₂(g) → CH₄(g)
ΔH°f(CH₄) = [ΔH°comb(C) + 2ΔH°comb(H₂)] – ΔH°comb(CH₄)
ΔH°f(CH₄) = [-393.5 + 2(-285.8)] – (-890.3) = -74.8 kJ/mol
Q.30 Complete the following reactions:
(a) CH₃-CH=CH₂ + H₂O + [O] (dil KMnO₄) / 273 K
CH₃-CH(OH)-CH₂(OH) (Propane-1,2-diol)
(b)
(c) H₃C-CH=CH₂ + HBr (Peroxide)
H₃C-CH₂-CH₂Br (1-Bromopropane) (Anti-Markovnikov addition)
SECTION – D
Q.31 (A) Define Common Ion Effect.
The common ion effect is the decrease in solubility of a salt when a soluble salt containing a common ion is added to the solution. Essentially, the presence of the common ion suppresses the ionization of the sparingly soluble salt.
(B) Describe the effect of: (any two)
(i) addition of H₂
Adding H₂ to the equilibrium 2H₂ + CO ⇌ CH₃OH will shift the equilibrium to the right, favoring the formation of CH₃OH (methanol).
(ii) removal of CO
Removing CO from the equilibrium will shift the equilibrium to the left, favoring the formation of H₂ and CO.
(iii) removal of CH₃OH
Removing CH₃OH will shift the equilibrium to the right, favoring the formation of CH₃OH.
(C) What is Kc for the following equilibrium when the equilibrium concentration of each substance is [SO₂] = 0.60 M, [O₂] = 0.82 M, and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]²[O₂]) = (1.90)² / ((0.60)² * 0.82) = 12.2
OR
(A) The equilibrium constant expression for a gas reaction is Kc = ([NO]²[H₂O]²)/([N₂][H₂]³). Write the balanced equation corresponding to this expression.
2N₂(g) + 3H₂(g) ⇌ 2NO(g) + 2H₂O(g)
(B) Find the value of Kc of the given equilibria from the value of Kp.
(1) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) Kp = 1.9 x 10⁻² at 500 K.
Kp = Kc(RT)^(Δn)
Δn = (2+1) – 2 = 1
R = 0.0821 L atm mol⁻¹ K⁻¹
Kc = Kp / (RT) = (1.9 x 10⁻²) / (0.0821 * 500) = 4.6 x 10⁻⁴
(C) Does the number of moles of reaction products increase, decrease, or remain the same when the following equilibria are subjected to a decrease in pressure by increasing the volume?
(i) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Increasing the volume (decreasing pressure) will shift the equilibrium to the side with more moles of gas. In this case, both sides have 4 moles of gaseous species. So, there will be no change.
(ii) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
Increasing the volume will shift the equilibrium to the side with more moles of gas. In this case, the left side has 1 mole of gas (CO₂) and the right side has none. So, the reaction will shift to the left (reactants).
Q.32 (A) Which among the following pair represents the correct IUPAC name for the compounds concerned?
(i) 2-chloro-4-methylpentane OR 4-chloro-2-methylpentane
Answer:4-chloro-2-methylpentane
(ii) But-3-yn-1-ol OR But-4-ol-1-yne
Answer:But-3-yn-1-ol
(B) Write IUPAC names of the products obtained by ozonolysis of 1-Phenylbut-1-ene.
Ozonolysis of 1-phenylbut-1-ene will yield:
- Benzaldehyde (C₆H₅CHO)
- Propanal (CH₃CH₂CHO)
(C) Draw the cis and trans structures for hex-2-ene.
CH3 H CH3 CH3
C=C C=C
H CH2CH2CH3 H CH2CH2CH3
cis-hex-2-ene trans-hex-2-ene
OR
(A) Write IUPAC names of the products obtained by ozonolysis of 2-Ethylbut-1-ene.
Ozonolysis of 2-ethylbut-1-ene will yield:
- Propanal (CH₃CH₂CHO)
- Butanone (CH₃CH₂COCH₃)
(B) What are the necessary conditions for any system to be aromatic?
- Cyclic (ring-shaped) structure.
- Planar (flat) structure.
- Complete delocalization of π electrons in the ring.
- Hückel’s rule (4n+2) π electrons (where n is an integer).
(C) Arrange the given set of compounds in order of their decreasing relative reactivity with an electrophile:
Chlorobenzene > 2,4-dinitrochlorobenzene > p-nitrochlorobenzene
Explanation:
Electron-donating groups activate the benzene ring towards electrophilic attack, while electron-withdrawing groups deactivate it. Nitro groups are strongly deactivating.
Q.33 (A) Express the change in internal energy of a system when:
(i) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings.
ΔU = -q (since W=0 and heat is taken out, q is negative)
(ii) ‘w’ amount of work is done by the system and q amount of heat is supplied to the system.
ΔU = q – w
(B) Define standard enthalpy of fusion.
The standard enthalpy of fusion is the change in enthalpy when one mole of a solid substance is converted into a liquid at its melting point under standard conditions (usually 1 atm pressure).
(C) Which among the following are intensive properties of a system? enthalpy, pressure, internal energy, density, volume
Intensive properties are independent of the amount of substance. Therefore:
- pressure
- density
(D) State the First Law of Thermodynamics.
Energy cannot be created or destroyed, only transferred between system and surroundings.
OR
(A) Define Entropy.
Entropy is a measure of the disorder or randomness of a system.
(B) For a reaction, both ΔH and ΔS are positive. Under what conditions will the reaction be spontaneous?
The reaction will be spontaneous when ΔG is negative. Since ΔG = ΔH – TΔS, and both ΔH and ΔS are positive, the reaction will be spontaneous at high temperatures.
(C) An exothermic reaction A → B is spontaneous in the backward direction. What will be the sign of ΔS for the forward reaction?
If the backward reaction is spontaneous, it means ΔG for the backward reaction is negative. Since ΔG = ΔH – TΔS, and the reaction is exothermic (ΔH is negative), the backward reaction is favored by enthalpy (negative ΔH). For the reaction to be spontaneous in the reverse direction, the entropy must be decreasing (ΔS < 0) for the reverse reaction. Therefore, for the forward reaction, ΔS will be positive (ΔS > 0).
(D) Enthalpy of combustion of carbon to CO₂ is -393.5 kJ/mol. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Moles of CO₂ = mass / molar mass = 35.2 g / 44 g/mol = 0.8 mol
Heat released = moles of CO₂ * enthalpy of combustion = 0.8 mol * 393.5 kJ/mol = 314.8 kJ