One mole each of CaC2, Al4C. and Mg2C3 reacts with H2O in separate open flasks at 25°C. Numerical value of the work done by the system is in order.

W=−pV2−V1=−pnRTP−0=−nRTInitial volume = 0 (being solid)Final volume = due to n moles of gas from I mole reactantCaC1mol(s)+2H2O⟶Ca(OH)2+C2H2(g)Al4C3(s)+12H2O⟶4Al(OH)3+3CH4Mg2C3 (s)+H2O⟶2Mg(OH)2+CH3−C1 mol≡CHThus, moles of C2H2=CH3−C≡CH<CH4Hence, numerical value of the work done by the system is in order.CaC2=Mg2C3,Al4C3

One mole each of CaC2, Al4C. and Mg2C3 reacts with H2O in separate open flasks at 25°C. Numerical value of the work done by the system is in order.

W=−pV2−V1=−pnRTP−0=−nRTInitial volume = 0 (being solid)Final volume = due to n moles of gas from I mole reactantCaC1mol(s)+2H2O⟶Ca(OH)2+C2H2(g)Al4C3(s)+12H2O⟶4Al(OH)3+3CH4Mg2C3 (s)+H2O⟶2Mg(OH)2+CH3−C1 mol≡CHThus, moles of C2H2=CH3−C≡CH<CH4Hence, numerical value of the work done by the system is in order.CaC2=Mg2C3,Al4C3

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One mole each of CaC₂, Al₄C. and Mg₂C₃ reacts with H₂O in separate open flasks at 25°C. Numerical value of the work done by the system is in order.

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${CaC}_{2} < {Al}_{4} C_{3} < {Mg}_{2} C_{3}$

${CaC}_{2} = {Mg}_{2} C_{3} < {Al}_{4} C_{3}$

${CaC}_{2} = {Mg}_{2} C_{3} = {Al}_{4} C_{3}$

${CaC}_{2} < {Al}_{4} C_{3} = {Mg}_{2} C_{3}$

answer is B.

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Detailed Solution

$W = - p (V_{2} - V_{1}) = - p (\frac{nRT}{P} - 0) = - nRT$

Initial volume = 0 (being solid)

Final volume = due to n moles of gas from I mole reactant

$\begin{array}{l} {CaC}_{1 mol} (s) + 2 H_{2} O ⟶ Ca (OH)_{2} + C_{2} H_{2} (g) \\ {Al}_{4} C_{3} (s) + 12 H_{2} O ⟶ 4 Al (OH)_{3} + 3 {CH}_{4} \\ \underset{}{{Mg}_{2} C_{3}} (s) + H_{2} O ⟶ 2 Mg (OH)_{2} + {CH}_{3} - \underset{1 mol}{C} \equiv CH \end{array}$