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Question
Calculate the work done in the decomposition of 132 g of \[\ce{NH4NO3}\] at 100°C.
\[\ce{NH4NO3_{(s)} -> N2O_{(g)} + 2H2O_{(g)}}\]
State whether work is done on or by the system.
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Solution
Given: Decomposition of 1 mole of NH4NO3
Temperature = T = 100°C = 373 K
To find: Work done and to determine whether work is done on the system or by the system.
Formula: W = −Δ ngRT
Calculation:
Molar mass of NH4NO3 = (2 × 14) + (3 × 16) + (4 × 1) = 80 g mol−1
Moles of NH4NO3 = n = `(132 "g")/(80 "g mol"^-1)` = 1.65 mol
The given reaction is for 1 mole of NH4NO3. For 1.65 moles of NH4NO3, the reaction is given as follows:
\[\ce{1.65 NH4NO3_{(s)} -> 1.65 N2O_{(g)} + 3.30 H2O_{(g)}}\]
Now,
Δng = (moles of product gases) − (moles of reactant gases)
Δng = (1.65 + 3.30) − 0 = +4.95 mol ...(∵ NH4NO3 is in solid state)
Hence,
W = −Δng RT
= − (+ 4.95 mol) × 8.314 J K−1 mol−1 × 373 K
= −15350 J
= −15.35 kJ
Work is done by the system (since W < 0).
The work done is −15.35 kJ. The work is done by the system.
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