# Reaction-enthalpy for synthesis of 40 g hydrazine

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?

• For the 40g issue, consider that any energy values you calculate are "per mole", i.e. they refer to one mole of product. You have to take into account how many moles of hydrazine correspond to 40g of hydrazine. – MarcoB Jun 6 '15 at 17:55

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 $ΔH_{rxn}$.