Given the following standard reaction enthalpies:
O₂(g) $\rightarrow$ 2O(g)
$\Delta$ H° = +498.4 kJ.mol^-1
NO(g) + O₃(g) $\rightarrow$ NO₂(g) + O₂(g)
$\Delta$ H° = -200 kJ.mol^-1
And the standard enthalpy of formation of ozone,+142.7 kJ.mol^-1,calculate the standard reaction enthalpy for the reaction
NO(g) + O(g) $\rightarrow$ NO₂(g)

Question

Given the following standard reaction enthalpies:
O₂(g)

\rightarrow

2O(g)

\Delta

H° = +498.4 kJ.mol^-1
NO(g) + O₃(g)

\rightarrow

NO₂(g) + O₂(g)

\Delta

H° = -200 kJ.mol^-1
And the standard enthalpy of formation of ozone,+142.7 kJ.mol^-1,calculate the standard reaction enthalpy for the reaction
NO(g) + O(g)

\rightarrow

NO₂(g)

Quizplus · Accepted Answer

The correct answer is A. The standard reaction enthalpy for the reaction NO(g)+ O(g)→NO2(g) can be calculated using Hess's law:NO(g)+ O3(g)→NO2(g)+ O2(g) ΔH° = -200 kJ.mol-1O2(g)→2O(g) ΔH° = +498.4 kJ.mol-1Adding these two reactions gives:NO(g)+ O(g)→NO2(g) ΔH° = -200 kJ.mol-1 + 498.4 kJ.mol-1 = +298.4 kJ.mol-1However, we need to take into account the standard enthalpy of formation of ozone, which is +142.7 kJ.mol-1. This means that we need to subtract this value from the ΔH° for the first reaction:NO(g)+ O3(g)→NO2(g)+ O2(g) ΔH° = -200 kJ.mol-1 - 142.7 kJ.mol-1 = -342.7 kJ.mol-1Adding this reaction to the previous one gives:NO(g)+ O(g)→NO2(g) ΔH° = +298.4 kJ.mol-1 - 342.7 kJ.mol-1 = -306 kJ.mol-1Therefore, the standard reaction enthalpy for the reaction NO(g)+ O(g)→NO2(g) is -306 kJ.mol-1.

Given the Following Standard Reaction Enthalpies: O2(g) →\rightarrow→ 2O(g) Δ\DeltaΔ H° = +4984 KJ

Given the Following Standard Reaction Enthalpies:
O₂(g) $\rightarrow$ 2O(g) $\Delta$ H° = +4984 KJ