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Answer the following questions · Q14

Q.xiv. Calculate the amount of work done in the

(a) oxidation of 1 mole HCl(g) at 200 0^0C according to reaction. 4HCl(g)+O2(g)→2 Cl2(g)+2 H2O(g)\mathrm{4HCl(g) + O_2(g) \rightarrow 2\,Cl_2(g) + 2\,H_2O(g)}
(b) decomposition of one mole of NO at 300 0^0C for the reaction 2 NO(g)→N2(g)+O2\mathrm{2\,NO(g) \rightarrow N_2(g) + O_2} Ans. : (a = + 983 kJ ; b = 0 kJ)
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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Step 1 (a). For 4HCl(g)+O2(g)→2Cl2(g)+2H2O(g) as written, gas reactant moles = 4+1=5, gas product moles=2+2=4, so Δng=4-5=-1 mol PER the equation as written (per 4 mol HCl). For 1 mole of HCl reacting, scale by 1/4: Δng = -1/4 mol.

Step 2. T = 200 degC = 473 K. W = -Δng RT = -(-1/4)(8.314)(473) = 0.25 x 8.314 x 473 = 983.1 J.

Step 3. FIDELITY NOTE: The source page prints 'Ans. : (a = +983 kJ)', but the correctly-scaled calculation for 1 mole of HCl gives 983 J (0.983 kJ), not 983 kJ -- the magnitude 983 matches exactly, so the printed 'kJ' unit appears to be a units typo for 'J' in the source, not a genuine discrepancy in the numeric working. …

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