21. calculate the $delta h$ for the reaction $ch_{4}(g)+nh_{3}(g)\rightarrow hcn(g)+3h_{2}(g)$ from the…

21. calculate the $delta h$ for the reaction $ch_{4}(g)+nh_{3}(g)\rightarrow hcn(g)+3h_{2}(g)$ from the following data: (8 pts) $n_{2}(g)+3h_{2}(g)\rightarrow 2nh_{3}(g)$ $delta h=-91.8 kj/mole$ $c(s)+2h_{2}(g)\rightarrow ch_{4}(g)$ $delta h=-74.9 kj/mole$ $h_{2}(g)+2c(s)+n_{2}(g)\rightarrow 2hcn(g)$ $delta h = + 270.3 kj/mole$

21. calculate the $delta h$ for the reaction $ch_{4}(g)+nh_{3}(g)\rightarrow hcn(g)+3h_{2}(g)$ from the following data: (8 pts) $n_{2}(g)+3h_{2}(g)\rightarrow 2nh_{3}(g)$ $delta h=-91.8 kj/mole$ $c(s)+2h_{2}(g)\rightarrow ch_{4}(g)$ $delta h=-74.9 kj/mole$ $h_{2}(g)+2c(s)+n_{2}(g)\rightarrow 2hcn(g)$ $delta h = + 270.3 kj/mole$

Answer

Explanation:

Step1: Reverse and scale reactions

Reverse the first reaction: $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$, $\Delta H = + 91.8\ kJ/mol$. Reverse the second reaction: $CH_4(g)\rightleftharpoons C(s)+2H_2(g)$, $\Delta H=+74.9\ kJ/mol$. Scale the third reaction by $\frac{1}{2}$: $\frac{1}{2}H_2(g)+C(s)+\frac{1}{2}N_2(g)\rightleftharpoons HCN(g)$, $\Delta H = + 135.15\ kJ/mol$.

Step2: Add the adjusted reactions

Add the three adjusted - reactions together: [ \begin{align*} 2NH_3(g)&\rightleftharpoons N_2(g)+3H_2(g)&\Delta H_1&= + 91.8\ kJ/mol\ CH_4(g)&\rightleftharpoons C(s)+2H_2(g)&\Delta H_2&=+74.9\ kJ/mol\ \frac{1}{2}H_2(g)+C(s)+\frac{1}{2}N_2(g)&\rightleftharpoons HCN(g)&\Delta H_3&= + 135.15\ kJ/mol \end{align*} ] [CH_4(g)+NH_3(g)\rightleftharpoons HCN(g)+3H_2(g)]

Step3: Calculate $\Delta H$

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$. $\Delta H=(91.8 + 74.9+135.15)\ kJ/mol$. $\Delta H = 301.85\ kJ/mol$.

Answer:

$301.85\ kJ/mol$