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Question

One mole each of CaC2, Al4C3 and Mg2C3 react with H2O in separate open flasks at 25oC. Numerical value of the work done by the system is in the order: 
Given reactions:
CaC2(s)+2H2O(l)Ca(OH)2(s)+C2H2(g)
Al4C3(s)+12H2O(l)4Al(OH)3(s)+3CH4(g)
Mg2C3(s)+4H2O(l)2Mg(OH)2(s)+CH3CCH(g)

 
  1. CaC2<Al4C3<Mg2C3
  2. CaC2=Mg2C3<Al4C3
  3. CaC2<Mg2C3<Al4C3
  4. CaC2>Al4C3>Mg2C3


Solution

The correct option is B CaC2=Mg2C3<Al4C3
Work done (W) = pV
W=p(V2V1)=p(nRTP0)=nRT

Initial volume = 0 (as no gaseous reactant are there)
Final volume = due to n moles of gas produced from 1 mole of reactant. 

CaC2(s)1 mol+2H2O(l)Ca(OH)2(s)+C2H2(g)1 mol
n (moles of gas in product) = 1

Al4C3(s)1 mol+12H2O(l)4Al(OH)3(s)+3CH43 mol(g)
n = 3

Mg2C3(s)1 mol+H2O(l)2Mg(OH)2(s)+CH3CCH1 mol(g)
n = 1

Thus, moles of (C2H2)=(CH3CCH)<CH4

The more the change in the number of moles, more will be the numerical work done in the reaction.

Hence, the numerical value of the work done by the system is in the order: CaC2=Mg2C3<Al4C3

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