One mole each of $C a C_{2}, A l_{4} C_{3} a n d M g_{2} C_{3}$ react with $H_{2} O$ in separate open flasks at $25^{o} C .$ Numerical value of the work done by the system is in the order:
Given reactions:
$C a C_{2} (s) + 2 H_{2} O (l) \to C a (O H)_{2} (s) + C_{2} H_{2} (g)$
$A l_{4} C_{3} (s) + 12 H_{2} O (l) \to 4 A l (O H)_{3} (s) + 3 C H_{4} (g)$
$M g_{2} C_{3} (s) + 4 H_{2} O (l) \to 2 M g (O H)_{2} (s) + C H_{3} - C \equiv C H (g)$

C a C_{2} < A l_{4} C_{3} < M g_{2} C_{3}

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C a C_{2} = M g_{2} C_{3} < A l_{4} C_{3}

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C a C_{2} < M g_{2} C_{3} < A l_{4} C_{3}

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C a C_{2} > A l_{4} C_{3} > M g_{2} C_{3}

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Solution

The correct option is B $C a C_{2} = M g_{2} C_{3} < A l_{4} C_{3}$
Work done (W) = $- p △ V$
$W = - p (V_{2} - V_{1}) = - p (\frac{n R T}{P} - 0) = - n R T$

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

$C a C_{2} (s) 1 m o l + 2 H_{2} O (l) \to C a (O H)_{2} (s) + C_{2} H_{2} (g) 1 m o l$
n (moles of gas in product) = 1

$A l_{4} C_{3} (s) 1 m o l + 12 H_{2} O (l) \to 4 A l (O H)_{3} (s) + 3 C H_{4} 3 m o l (g)$
n = 3

$M g_{2} C_{3} (s) 1 m o l + H_{2} O (l) \to 2 M g (O H)_{2} (s) + C H_{3} - C \equiv C H 1 m o l (g)$
n = 1

Thus, moles of $(C_{2} H_{2}) = (C H_{3} - C \equiv C H) < C H_{4}$

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: $C a C_{2} = M g_{2} C_{3} < A l_{4} C_{3}$

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Similar questions

C a C_{2}, A l_{4} C_{3} a n d M g_{2} C_{3}

H_{2} O

25^{o} C .

C a C_{2} (s) + 2 H_{2} O (l) \to C a (O H)_{2} (s) + C_{2} H_{2} (g)

A l_{4} C_{3} (s) + 12 H_{2} O (l) \to 4 A l (O H)_{3} (s) + 3 C H_{4} (g)

M g_{2} C_{3} (s) + 4 H_{2} O (l) \to 2 M g (O H)_{2} (s) + C H_{3} - C \equiv C H (g)

Which is not obtained when metal carbides react with H₂O?

B e_{2} C, C a C_{2}, M g_{2} C_{3}, A l_{4} C_{3}, C H_{4},

C a C_{2}

A l_{4} C_{3}

S i C, C a C_{2}, B e_{2} C, A l_{2} C_{6}, A l_{4} C_{3}, B_{4} C

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