For the gaseous reaction, $2 ext{CO}(g) + ext{O}_2(g) ightarrow 2 ext{CO}_2(g)$; $\Delta H = -

Question

For the gaseous reaction, $2	ext{CO}(g) + 	ext{O}_2(g) ightarrow 2	ext{CO}_2(g)$; $\Delta H = -560	ext{ kJ}$. Two moles of $	ext{CO}$ and one mole of $	ext{O}_2$ are taken in a closed rigid container of volume $1	ext{ L}$. They completely react to form two moles of $	ext{CO}_2$, and the gases deviate appreciably from ideal behavior. If the pressure in the vessel changes from $70	ext{ atm}$ to $40	ext{ atm}$, find the magnitude (absolute value) of the change in internal energy, $|\Delta U|$ (in $	ext{kJ}$), at $500	ext{ K}$. (Take $1	ext{ L atm} = 0.1	ext{ kJ}$)

Professor PC Thomas Classes · Answer

By definition of enthalpy for any general process: $\Delta H = \Delta U + \Delta(PV) = \Delta U + (P_2V_2 - P_1V_1)$. Since the container is rigid, $V_1 = V_2 = V = 1	ext{ L}$. Thus, $\Delta(PV) = V(P_2 - P_1) = 1	ext{ L} 	imes (40 - 70)	ext{ atm} = -30	ext{ L atm}$. Using the conversion factor $1	ext{ L atm} = 0.1	ext{ kJ}$, we get $\Delta(PV) = -30 	imes 0.1 = -3	ext{ kJ}$. Substituting this into the enthalpy equation gives $-560 = \Delta U - 3 \implies \Delta U = -557	ext{ kJ}$. Therefore, the magnitude $|\Delta U| = 557	ext{ kJ}$.