## Chemistry (4th Edition)

c) w= $-5.4\times10^{2}$ kJ, done by the system.
q = -928 kJ. (Heat is released, so q is negative.) $\Delta U = -1.47\times10^{3} kJ.$ $\Delta U$ = q+w, Or w=$\Delta U-q = -1.47\times10^{2} kJ -(-928 kJ)$ = $-5.4\times10^{2}$ kJ. Negative sign indicates that the work is done by the system.