Determine $H$, $Q$, $W$, and $E$ at $298\ \mathrm K$ and $1\ \mathrm{atm}$ for the complete reaction of $8.170\ \mathrm g$ of $\ce{N2O4}$.

$$\ce{N2O4(g) + 2N2H4(l) -> 3N2(g) + 4H2O(l)}$$

The heat of formation of liquid hydrazine ($\ce{N2H4}$) is $50.63\ \mathrm{kJ/mol}$. Use tabulated data for other heats of formation.

The tolerance on each question is only $0.03\ \mathrm{kJ}$, so express all answers to $0.01\ \mathrm{kJ}$.

Hello, I am having a lot of difficulty trying to do this problem. I found the moles of $\ce{N2O4}$ and got $0.0888$ moles. I got $0.17761$ moles of hydrazine and $0.3552$ moles of water. I found $\Delta H$ by doing $(0.3552)(-2.85.8)-(0.0888)(9.66)+(0.17761)(50.63)$ and got $-93.381$ which is wrong. Also $\Delta H$ and $Q$ should be the same solution right? For $W$ I did $nRT$ $(0.17761)(8.314)(298)= 440.04$.


There is a mistake in the calculation of the heat of reaction. It must be :

$\Delta H = 0.355·(-286)kJ - (0.1776·50.63)kJ - (0.0888·9.7)kJ = - 101.53 kJ - 9.00 kJ - 0.86 kJ = - 111.40 kJ$

Also the work is not nRT. It is $\Delta n$ RT. And $\Delta n$ is twice the number of moles of $N_2O_4$

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