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Answer the following questions · Q8

Q.viii. Calculate the work done in the decomposition of 132 g of NH4_4NO3_3 at 100 0^0C. NH4NO3(s)→N2O(g)+2 H2O(g)\mathrm{NH_4NO_3(s) \rightarrow N_2O(g) + 2\,H_2O(g)} State whether work is done on the system or by the system. Ans. : (-18.6 kJ)

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Step 1. Molar mass NH4NO3 = 14+4(1)+14+3(16) = 80 g/mol. Moles of NH4NO3 = 132 g / 80 g mol-1 = 1.65 mol.

Step 2. For NH4NO3(s) → N2O(g) + 2H2O(g), each mole of solid reactant produces 1+2 = 3 moles of gas from 0 moles of gas reactant, so Δng = +3 per mole reacted.

Step 3. For 1.65 mol reacting: Δng(total) = 1.65 x 3 = 4.95 mol. T = 100 degC = 373 K.

Step 4. W = -Δng RT = -4.95 mol x 8.314 J K-1 mol-1 x 373 K = -15,350 J = -15.35 kJ. The negative sign means work is done BY the system (the solid decomposing into 3x its own moles of gas pushes back the surroundings). …

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