Question

Given the reaction: $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)+22kcal$.
When $6$ moles of $N{H}_3$ are consumed to produce nitrogen gas and hydrogen gas, determine the value of $\Delta H$ for the reverse reaction?

• A. $+22kcal$
• B. $+66kcal$
• C. $-22kcal$
• D. $-66kcal$
• E. $+33kcal$

Answer 1

The correct option is B $+66kcal$

Given $N_2$ + $3H_2$ ⇌ $2N{H}_3$ (Exothermic Reaction) $\Delta H=-22kcal$

6 moles of $N{H}_3$ are produced. Reverse the reaction

$2N{H}_3$ + $22kcal$ ⇌ $N_2$ + $3H_2$ $\Delta H$ = +22 kcal ( Reaction is Endothermic when we reverse the Reaction)

Multiply the Reaction with 3

$6N{H}_3$ + 66kcal ⇌ $N_2$ + $3H_2$

So, $\Delta H=$ +66 kcal