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

To calculate the standard reaction enthalpy for the given reaction, we need to use Hess's Law, which states that the overall enthalpy change of a reaction is equal to the sum of enthalpy changes of its individual steps. First, we need to balance the equation: 2NO(g) + O2(g) -> 2NO2(g) Now, we can use the given standard enthalpies to calculate the enthalpy change for this reaction: Enthalpy change for the first step: NO(g) + O₃(g) -> NO₂(g) + O₂(g) = F1F1F1S1−F1F1F10200 kJ Enthalpy change for the second step: NO₂(g) + O2(g) -> 2NO2(g) = F1F1F1S1−F1F1F10306 kJ Adding the two enthalpy changes gives: Enthalpy change for the overall reaction: NO(g) + O2(g) -> 2NO2(g) = F1F1F1S1−F1F1F10200 kJ + F1F1F1S1−F1F1F10306 kJ = F1F1F1S1−F1F1F10306 kJ (as the F1F1F1S1−F1F1F10200 kJ term cancels out) Therefore, the correct answer is A.

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