Chemistry Q130 – AP-EAMCET 2025

Home Chemistry Chemical ThermodynamicsQuestion #296

The value of

Δ_{r} H

for the reaction

N_{2} O_{4} (g) + 3 CO(g) ⟶ N_{2} O(g) + 3 {CO}_{2} (g)

(in kJ) is (Given : Enthalpies of formation of

CO(g)

{CO}_{2} (g)

N_{2} O(g)

and

N_{2} O_{4} (g)

are respectively

- 110

- 393

81

and

9.7 {kJ mol}^{- 1}

)

Clue 1 of 3: Hess's Law

The enthalpy change of a reaction can be calculated from the standard enthalpies of formation of the reactants and products.

🧠 Full Solution Path

Step 1: Identify Reaction and Given Data✦ Active

The given chemical reaction is: $N_{2} O_{4} (g) + 3 CO(g) ⟶ N_{2} O(g) + 3 {CO}_{2} (g)$ The standard enthalpies of formation ( $Δ_{f} H^{\circ}$ ) for the species involved are:

Δ_{f} H^{\circ} (CO(g)) = - 110 {kJ mol}^{- 1}

Δ_{f} H^{\circ} ({CO}_{2} (g)) = - 393 {kJ mol}^{- 1}

Δ_{f} H^{\circ} (N_{2} O(g)) = 81 {kJ mol}^{- 1}

Δ_{f} H^{\circ} (N_{2} O_{4} (g)) = 9.7 {kJ mol}^{- 1}

💡 Teacher's Secret Hint

Pay close attention to the sign (positive or negative) of each enthalpy of formation value, as this is a common source of error in calculations.

Step 2: Calculate Enthalpy Sum for Products○ Expand

The products are $N_{2} O(g)$ and ${CO}_{2} (g)$ . Their stoichiometric coefficients are $1$ and $3$ , respectively.

\sum n_{p} Δ_{f} H^{\circ} (products) = [1 \times Δ_{f} H^{\circ} (N_{2} O(g))] + [3 \times Δ_{f} H^{\circ} ({CO}_{2} (g))]

= [1 \times (81 {kJ mol}^{- 1})] + [3 \times (- 393 {kJ mol}^{- 1})]

= 81 kJ - 1179 kJ

= - 1098 kJ

💡 Teacher's Secret Hint

Remember to multiply each enthalpy of formation by its corresponding stoichiometric coefficient from the balanced chemical equation.

Step 3: Calculate Enthalpy Sum for Reactants○ Expand

The reactants are $N_{2} O_{4} (g)$ and $CO(g)$ . Their stoichiometric coefficients are $1$ and $3$ , respectively.

\sum n_{r} Δ_{f} H^{\circ} (reactants) = [1 \times Δ_{f} H^{\circ} (N_{2} O_{4} (g))] + [3 \times Δ_{f} H^{\circ} (CO(g))]

= [1 \times (9.7 {kJ mol}^{- 1})] + [3 \times (- 110 {kJ mol}^{- 1})]

= 9.7 kJ - 330 kJ

= - 320.3 kJ

💡 Teacher's Secret Hint

Double-check the signs of the enthalpy values, especially for compounds like $CO(g)$ which have negative formation enthalpies.

Step 4: Determine the Standard Enthalpy of Reaction○ Expand

The standard enthalpy of reaction, $Δ_{r} H^{\circ}$ , is calculated using the formula:

Δ_{r} H^{\circ} = \sum n_{p} Δ_{f} H^{\circ} (products) - \sum n_{r} Δ_{f} H^{\circ} (reactants)

= (- 1098 kJ) - (- 320.3 kJ)

= - 1098 kJ + 320.3 kJ

= - 777.7 kJ

Rounding to the nearest integer, the value is $- 778 kJ$ .

💡 Teacher's Secret Hint

Be careful with the subtraction of negative numbers. A common mistake is to forget to change the sign when subtracting a negative value.

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Chemistry Question 130 – AP-EAMCET 2025

Clue 1 of 3: Hess's Law

🧠 Full Solution Path