Question

The reaction of nitrogen with hydrogen to make ammonia has $\Delta H=-92kJ$ $N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$. What is the value of $\Delta U$ (in kJ) if the reaction is carried out at a constant pressure of $40bar$ and the volume change is $-1.25litre$?

Answer 1

$\Delta H=\Delta U+P\Delta V$

$\Delta U=\Delta H-P\Delta V$

$\Delta U=-92\times {10}^3-(40\times 1.013\times -1.25)$ [Since, $1atm=1.013bar,P=40\times 1.013atm$]

$\Delta U=-92,000+50.65=-91,949J=-91.949\times {10}^{3}J$

$\Delta U=-91.949kJ$