Given the following reactions N₂ (g) + O₂ (g) → 2NO (g) ΔH = +180.7 kJ
2NO (g) + O₂ (g) → 2NO₂ (g) ΔH = -113.1 kJ
The enthalpy of reaction for
4NO (g) → 2NO₂ (g) + N₂ (g)
Is ________ kJ.

Question

Quizplus · Accepted Answer

We need to reverse the second equation and multiply it by 2 so that the products match the reactants in the third equation. Then we need to reverse the first equation and add it to the second equation multiplied by 4 to cancel out the NO and obtain the desired reaction. This gives us: 2NO₂ (g) → 2NO (g) + O₂ (g)ΔH = +113.1 kJ (reverse of second equation) 4NO (g) + 2O₂ (g) → 2NO₂ (g) + 2N₂ (g) (multiply second equation by 4 and reverse first equation) Now we can sum up the enthalpies of reaction for these two equations to obtain the enthalpy of reaction for the desired equation: 4NO (g) → 2NO₂ (g) + N₂ (g)ΔH = -293.8 kJ

Given the Following Reactions N2 (G) + O2 (G) →

Given the Following Reactions N₂ (G) + O₂ (G) →