# Determine H, q, w, and E at 298 K and 1 atm for the complete reaction of 8.170 g of N2O4.?

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 $$nRT$$ $$(0.17761)(8.314)(298)= 440.04$$.

$$\Delta H = 0.355·(-286)kJ - (0.1776·50.63)kJ - (0.0888·9.7)kJ = - 101.53 kJ - 9.00 kJ - 0.86 kJ = - 111.40 kJ$$
Also the work is not nRT. It is $$\Delta n$$ RT. And $$\Delta n$$ is twice the number of moles of $$N_2O_4$$