Given the enthalpies of formation: CO(g) = -110 kJ mol⁻¹, CO₂(g) = -393 kJ mol⁻¹, N₂O(g) = 81 kJ mol

Q: What is the correct answer for CBSE11?

The correct answer is C. This is Exercise 5.12. Using Δ_r H° = Σa_iΔ_f H°(products) - Σb_iΔ_f H°(reactants): Δ_r H° = [1 * Δ_f H°(N₂O,g) + 3 * Δ_f H°(CO₂,g)] - [1 * Δ_f H°(N₂O₄,g) + 3 * Δ_f H°(CO,g)] Δ_r H° = [1 * (81) + 3 * (-393)] - [1 * (9.7) + 3 * (-110)] Δ_r H° = [81 - 1179] - [9.7 - 330] Δ_r H° = [-1098] - [-320.3] Δ_r H° = -1098 + 320.3 = -777.7 kJ mol⁻¹. Let's recheck the values from the table. CO(g) = -110.53, CO

Q: Given the enthalpies of formation: CO(g) = -110 kJ mol⁻¹, CO₂(g) = -393 kJ mol⁻¹, N₂O(g) = 81 kJ mol⁻¹, N₂O₄(g) = 9.7 kJ mol⁻¹. Calculate Δ_r H for the reaction: N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g).

The correct answer is C. This is Exercise 5.12. Using Δ_r H° = Σa_iΔ_f H°(products) - Σb_iΔ_f H°(reactants): Δ_r H° = [1 * Δ_f H°(N₂O,g) + 3 * Δ_f H°(CO₂,g)] - [1 * Δ_f H°(N₂O₄,g) + 3 * Δ_f H°(CO,g)] Δ_r H° = [1 * (81) + 3 * (-393)] - [1 * (9.7) + 3 * (-110)] Δ_r H° = [81 - 1179] - [9.7 - 330] Δ_r H° = [-1098] - [-320.3] Δ_r H° = -1098 + 320.3 = -777.7 kJ mol⁻¹. Let's recheck the values from the table. CO(g) = -110.53, CO

Q: What are the options for this CBSE11 question?

A. -777.3 kJ mol⁻¹ · B. -747.3 kJ mol⁻¹ · C. -717.3 kJ mol⁻¹ · D. -687.3 kJ mol⁻¹

Question

Given the enthalpies of formation: CO(g) = -110 kJ mol⁻¹, CO₂(g) = -393 kJ mol⁻¹, N₂O(g) = 81 kJ mol⁻¹, N₂O₄(g) = 9.7 kJ mol⁻¹. Calculate Δ_r H for the reaction: N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g).

FirstInTest · Accepted Answer

This is Exercise 5.12. Using Δ_r H° = Σa_iΔ_f H°(products) - Σb_iΔ_f H°(reactants):
Δ_r H° = [1 * Δ_f H°(N₂O,g) + 3 * Δ_f H°(CO₂,g)] - [1 * Δ_f H°(N₂O₄,g) + 3 * Δ_f H°(CO,g)]
Δ_r H° = [1 * (81) + 3 * (-393)] - [1 * (9.7) + 3 * (-110)]
Δ_r H° = [81 - 1179] - [9.7 - 330]
Δ_r H° = [-1098] - [-320.3]
Δ_r H° = -1098 + 320.3 = -777.7 kJ mol⁻¹. Let's recheck the values from the table. CO(g) = -110.53, CO2(g) = -393.51, N2O(g) = 81.6, N2O4(g) = 9.16. Using these values:
Δ_r H° = [1 * (81.6) + 3 * (-393.51)] - [1 * (9.16) + 3 * (-110.53)]
Δ_r H° = [81.6 - 1180.53] - [9.16 - 331.59]
Δ_r H° = [-1098.93] - [-322.43]
Δ_r H° = -1098.93 + 322.43 = -776.5 kJ mol⁻¹. The options are rounded. The closest is -777.3 kJ mol⁻¹. The problem statement uses rounded values: -110, -393, 81, 9.7. Let's use those. Δ_r H° = [1 * (81) + 3 * (-393)] - [1 * (9.7) + 3 * (-110)] = [81 - 1179] - [9.7 - 330] = -1098 - (-320.3) = -1098 + 320.3 = -777.7 kJ mol⁻¹. So, the closest option is A. Let's recheck the options and my calculation. The question has -110, -393, 81, 9.7. My calculation gives -777.7. Option A is -777.3. This is a very close match, likely due to rounding in the options. I will select A.

Answer

-777.3 kJ mol⁻¹

Answer

-747.3 kJ mol⁻¹

Answer

-717.3 kJ mol⁻¹

Answer

-687.3 kJ mol⁻¹

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