Question
Calculate the work done and comment on whether work is done on or by the system for the decomposition of $2$ moles of
$NH_4NO_3$ at $100 ^\circ C$
$NH_4NO_{3(s)} \rightarrow N_2O_{(g)} + 2H_2O_{(g)}$

Answer

Given : $NH_4NO_{3(s)} \rightarrow N_2O_{(g)} + 2H_2O_{(g)}$
$n_{NH4NO3} = 2 mol; T = 273 + 100 = 373 K$
W = ? Comment on work = ?
$\triangle n_{reaction} = (1 + 2) – 0 = 3 mol$
$\because For 1 mol of NH_4NO_3 \triangle n_{reaction} = 3 mol$
$\therefore For 2 mol of NH_4NO_3 \triangle n_{reaction} = 6 mol$
Due to $6$ moles of gaseous products from $2\ mol\  NH_4NO_3$, there is work of expansion,
​​​​​​​hence work is done by the system.
$W = -\triangle nRT$
$= – 6 \times 8.314 \times 373 = -18606 J$
$= -18.606 kJ$
Ans. Work is done by the system.
$W= -18.606\ kJ$

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