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Chemical Thermodynamics: Chemistry | JEE Main Enthalpies of formation of
C C l 4 ( g ) , H 2 O ( g ) , C O 2 ( g ) and H C l ( g ) are − 105 , − 242 , − 394 and − 92 k J m o l − 1 \mathrm{CCl}_4(\mathrm{~g}), \mathrm{H}_2\mathrm{O}(\mathrm{g}), \mathrm{CO}_2(\mathrm{~g})\text{and }\mathrm{HCl}(\mathrm{g})\text{are } -105, -242, -394\text{ and } -92\mathrm{~kJ}\mathrm{~mol}^{-1} CCl 4 ( g ) , H 2 O ( g ) , CO 2 ( g ) and HCl ( g ) are − 105 , − 242 , − 394 and − 92 kJ mol − 1 respectively. The magnitude of enthalpy of the reaction given below is
‾ k J m o l − 1 \underline{\quad\quad\quad} \mathrm{kJ}\mathrm{mol}^{-1} kJ mol − 1 . (nearest integer)
C C l 4 ( g ) + 2 H 2 O ( g ) → C O 2 ( g ) + 4 H C l ( g ) \mathrm{CCl}_4(\mathrm{~g})+2\mathrm{H}_2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+4\mathrm{HCl}(\mathrm{g}) CCl 4 ( g ) + 2 H 2 O ( g ) → CO 2 ( g ) + 4 HCl ( g ) Hint 1 of 3
Which formula correctly relates the standard enthalpy of reaction (Δ r H ∘ \Delta_r H^\circ Δ r H ∘ Δ f H ∘ \Delta_f H^\circ Δ f H ∘
Δ r H ∘ = ∑ n r Δ f H ∘ ( reactants ) − ∑ n p Δ f H ∘ ( products ) \Delta_r H^\circ = \sum n_r \Delta_f H^\circ(\text{reactants}) - \sum n_p \Delta_f H^\circ(\text{products}) Δ r H ∘ = ∑ n r Δ f H ∘ ( reactants ) − ∑ n p Δ f H ∘ ( products ) Δ r H ∘ = ∑ n p Δ f H ∘ ( products ) − ∑ n r Δ f H ∘ ( reactants ) \Delta_r H^\circ = \sum n_p \Delta_f H^\circ(\text{products}) - \sum n_r \Delta_f H^\circ(\text{reactants}) Δ r H ∘ = ∑ n p Δ f H ∘ ( products ) − ∑ n r Δ f H ∘ ( reactants ) No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.
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The magnitude of the enthalpy of the reaction is 173 kJ mol⁻¹. Option analysis
Why each option works or fails
Step Working
01 given Enthalpies of formation: Δ f H [ C C l 4 ( g ) ] = − 105 kJ mol − 1 \Delta_f H[\mathrm{CCl_4(g)}] = -105\text{ kJ mol}^{-1} Δ f H [ CC l 4 ( g ) ] = − 105 kJ mol − 1 Δ f H [ H 2 O ( g ) ] = − 242 kJ mol − 1 \Delta_f H[\mathrm{H_2O(g)}] = -242\text{ kJ mol}^{-1} Δ f H [ H 2 O ( g ) ] = − 242 kJ mol − 1 Δ f H [ C O 2 ( g ) ] = − 394 kJ mol − 1 \Delta_f H[\mathrm{CO_2(g)}] = -394\text{ kJ mol}^{-1} Δ f H [ C O 2 ( g ) ] = − 394 kJ mol − 1 Δ f H [ H C l ( g ) ] = − 92 kJ mol − 1 \Delta_f H[\mathrm{HCl(g)}] = -92\text{ kJ mol}^{-1} Δ f H [ HCl ( g ) ] = − 92 kJ mol − 1 C C l 4 ( g ) + 2 H 2 O ( g ) → C O 2 ( g ) + 4 H C l ( g ) \mathrm{CCl_4(g)} + 2\mathrm{H_2O(g)} \rightarrow \mathrm{CO_2(g)} + 4\mathrm{HCl(g)} CC l 4 ( g ) + 2 H 2 O ( g ) → C O 2 ( g ) + 4 HCl ( g )
02 find Magnitude of the enthalpy of the reaction (∣ Δ r H ∣ |\Delta_r H| ∣ Δ r H ∣ kJ mol − 1 \text{kJ mol}^{-1} kJ mol − 1
03 strategise Apply Hess's Law / enthalpy relation using enthalpies of formation:
Δ r H = ∑ n p Δ f H ( products ) − ∑ n r Δ f H ( reactants ) \Delta_r H = \sum n_p \Delta_f H(\text{products}) - \sum n_r \Delta_f H(\text{reactants}) Δ r H = ∑ n p Δ f H ( products ) − ∑ n r Δ f H ( reactants ) Δ r H = [ 1 × Δ f H ( C O 2 ) + 4 × Δ f H ( H C l ) ] − [ 1 × Δ f H ( C C l 4 ) + 2 × Δ f H ( H 2 O ) ] \Delta_r H = [1 \times \Delta_f H(\mathrm{CO_2}) + 4 \times \Delta_f H(\mathrm{HCl})] - [1 \times \Delta_f H(\mathrm{CCl_4}) + 2 \times \Delta_f H(\mathrm{H_2O})] Δ r H = [ 1 × Δ f H ( C O 2 ) + 4 × Δ f H ( HCl )] − [ 1 × Δ f H ( CC l 4 ) + 2 × Δ f H ( H 2 O )]
04 execute Substitute the given values:
Δ r H = [ ( − 394 ) + 4 ( − 92 ) ] − [ ( − 105 ) + 2 ( − 242 ) ] \Delta_r H = [(-394) + 4(-92)] - [(-105) + 2(-242)] Δ r H = [( − 394 ) + 4 ( − 92 )] − [( − 105 ) + 2 ( − 242 )] Δ r H = [ − 394 − 368 ] − [ − 105 − 484 ] \Delta_r H = [-394 - 368] - [-105 - 484] Δ r H = [ − 394 − 368 ] − [ − 105 − 484 ] Δ r H = − 762 − ( − 589 ) = − 762 + 589 = − 173 kJ mol − 1 \Delta_r H = -762 - (-589) = -762 + 589 = -173\text{ kJ mol}^{-1} Δ r H = − 762 − ( − 589 ) = − 762 + 589 = − 173 kJ mol − 1 ∣ Δ r H ∣ = ∣ − 173 ∣ = 173 kJ mol − 1 |\Delta_r H| = |-173| = 173\text{ kJ mol}^{-1} ∣ Δ r H ∣ = ∣ − 173∣ = 173 kJ mol − 1
✓ verify Exothermic reaction gives negative Δ r H = − 173 kJ mol − 1 \Delta_r H = -173\text{ kJ mol}^{-1} Δ r H = − 173 kJ mol − 1
Hints that build this answer step by step Which formula correctly relates the standard enthalpy of reaction (Δ r H ∘ \Delta_r H^\circ Δ r H ∘ Δ f H ∘ \Delta_f H^\circ Δ f H ∘
Δ r H ∘ = ∑ n p Δ f H ∘ ( products ) − ∑ n r Δ f H ∘ ( reactants ) \Delta_r H^\circ = \sum n_p \Delta_f H^\circ(\text{products}) - \sum n_r \Delta_f H^\circ(\text{reactants}) Δ r H ∘ = ∑ n p Δ f H ∘ ( products ) − ∑ n r Δ f H ∘ ( reactants ) Substituting the given values into the expression, what is the value of Δ r H ∘ \Delta_r H^\circ Δ r H ∘
[-394 + 4(-92)] - [-105 + 2(-242)] = -173 kJ mol⁻¹ The question asks for the magnitude of the enthalpy of the reaction. What is the final value?
173 Your next move We think you should solve this next ✓ Source and academic review↓
Question type Numerical
Exam relevance JEE Main · Chemistry
Concepts assessed Chemistry
Academic status Reviewed by official_key
Source pyq
Editorial review 9 September 2026 Quick checks
Students also ask Why is the answer 173 and not -173? The problem asks for the 'magnitude of enthalpy of the reaction', so the sign is omitted and only the absolute value is reported.
Answer The magnitude of the enthalpy of the reaction is 173 kJ mol⁻¹.
Why each option works or fails Step-by-step solution given: Enthalpies of formation: Δ f H [ C C l 4 ( g ) ] = − 105 kJ mol − 1 \Delta_f H[\mathrm{CCl_4(g)}] = -105\text{ kJ mol}^{-1} Δ f H [ CC l 4 ( g ) ] = − 105 kJ mol − 1 Δ f H [ H 2 O ( g ) ] = − 242 kJ mol − 1 \Delta_f H[\mathrm{H_2O(g)}] = -242\text{ kJ mol}^{-1} Δ f H [ H 2 O ( g ) ] = − 242 kJ mol − 1 Δ f H [ C O 2 ( g ) ] = − 394 kJ mol − 1 \Delta_f H[\mathrm{CO_2(g)}] = -394\text{ kJ mol}^{-1} Δ f H [ C O 2 ( g ) ] = − 394 kJ mol − 1 Δ f H [ H C l ( g ) ] = − 92 kJ mol − 1 \Delta_f H[\mathrm{HCl(g)}] = -92\text{ kJ mol}^{-1} Δ f H [ HCl ( g ) ] = − 92 kJ mol − 1 C C l 4 ( g ) + 2 H 2 O ( g ) → C O 2 ( g ) + 4 H C l ( g ) \mathrm{CCl_4(g)} + 2\mathrm{H_2O(g)} \rightarrow \mathrm{CO_2(g)} + 4\mathrm{HCl(g)} CC l 4 ( g ) + 2 H 2 O ( g ) → C O 2 ( g ) + 4 HCl ( g ) find: Magnitude of the enthalpy of the reaction (∣ Δ r H ∣ |\Delta_r H| ∣ Δ r H ∣ kJ mol − 1 \text{kJ mol}^{-1} kJ mol − 1 strategise: Apply Hess's Law / enthalpy relation using enthalpies of formation:
Δ r H = ∑ n p Δ f H ( products ) − ∑ n r Δ f H ( reactants ) \Delta_r H = \sum n_p \Delta_f H(\text{products}) - \sum n_r \Delta_f H(\text{reactants}) Δ r H = ∑ n p Δ f H ( products ) − ∑ n r Δ f H ( reactants ) Δ r H = [ 1 × Δ f H ( C O 2 ) + 4 × Δ f H ( H C l ) ] − [ 1 × Δ f H ( C C l 4 ) + 2 × Δ f H ( H 2 O ) ] \Delta_r H = [1 \times \Delta_f H(\mathrm{CO_2}) + 4 \times \Delta_f H(\mathrm{HCl})] - [1 \times \Delta_f H(\mathrm{CCl_4}) + 2 \times \Delta_f H(\mathrm{H_2O})] Δ r H = [ 1 × Δ f H ( C O 2 ) + 4 × Δ f H ( HCl )] − [ 1 × Δ f H ( CC l 4 ) + 2 × Δ f H ( H 2 O )] execute: Substitute the given values:
Δ r H = [ ( − 394 ) + 4 ( − 92 ) ] − [ ( − 105 ) + 2 ( − 242 ) ] \Delta_r H = [(-394) + 4(-92)] - [(-105) + 2(-242)] Δ r H = [( − 394 ) + 4 ( − 92 )] − [( − 105 ) + 2 ( − 242 )] Δ r H = [ − 394 − 368 ] − [ − 105 − 484 ] \Delta_r H = [-394 - 368] - [-105 - 484] Δ r H = [ − 394 − 368 ] − [ − 105 − 484 ] Δ r H = − 762 − ( − 589 ) = − 762 + 589 = − 173 kJ mol − 1 \Delta_r H = -762 - (-589) = -762 + 589 = -173\text{ kJ mol}^{-1} Δ r H = − 762 − ( − 589 ) = − 762 + 589 = − 173 kJ mol − 1 ∣ Δ r H ∣ = ∣ − 173 ∣ = 173 kJ mol − 1 |\Delta_r H| = |-173| = 173\text{ kJ mol}^{-1} ∣ Δ r H ∣ = ∣ − 173∣ = 173 kJ mol − 1 verify: Exothermic reaction gives negative Δ r H = − 173 kJ mol − 1 \Delta_r H = -173\text{ kJ mol}^{-1} Δ r H = − 173 kJ mol − 1