Reaction-enthalpy for synthesis of 40 g hydrazine

Question

Calculate the reaction-enthalpy for the synthesis of 40 g hydrazine (rocketfuel): $$\ce{4NH3(g) + Cl2 (g) -> N2H4(l) + 2 NH4Cl(s)}$$

Attempt at solution:

First we have to account for the decomposition of four moles of ammonia into its elements.

Since $\Delta H_\mathrm f^o(\ce{NH3})$ = $-46.1 \mathrm{~kJ~mol^{-1}}$, we do:

$$\Delta H_1 = 4(+46.1) = 184.4 \mathrm{~kJ}$$

$\ce{Cl2}$ is in its standard state so we can ignore that.

We know that $\Delta H_f^o(\ce{N2H4})$ = $50.6\mathrm{~kJ~mol^{-1}}$. Therefore, $\Delta H_2 = \pu{50.6 kJ}$.

Then, $\Delta H_\mathrm f^o(\ce{NH4Cl})$ = $-314.2\mathrm{~kJ~mol^{-1}}$ to give

$$\Delta H_3 = 2(-314.4) = -628.8~\mathrm{kJ}$$

Finally, \begin{align} \Delta H_{\mathrm{rxn}} &= \Delta H_1 + \Delta H_2 + \Delta H_3\\ &= 184.4 + 50.6 + (-628.8) = -393.8~\mathrm{kJ} \end{align}

However, my textbook says the correct answer is $-492.3~\mathrm{kJ}$.

Can someone tell me what I did wrong, please? Also, I didn't account for the $40\mathrm{~g}$. How does that affect the problem?

Answer 1

You are on the right track. Basically, you are just missing the last step. Your answer for the enthalpy of reaction should be $ΔH_{rxn}=−393.8\frac{\mathrm{kJ}}{\mathrm{mol}}$. So all you have to do is find the moles of hydrazine
$n=\frac{40}{32.0452}=1.248\space\mathrm{moles}$ and multiply it by your $ΔH_{rxn}$.