$2$ mole of zinc is dissolved in $H C l$ at $25^{o} C$ . The work done in open vessel is:

- 2.477 k J

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- 4.955 k J

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0.0489 k J

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None of these

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Solution

The correct option is D $- 4.955 k J$
$W$ = $Δ n_{g} R T$

Given: $T = 25^{o} C = 298 K$
$R = 8.314$

$Z n + 2 H C l ⟶ Z n C l_{2} + H_{2}$

$Δ n_{g} = 1 - 0 = 1$

$W = - 1 \times 8.314 \times 298$

For 2 moles,

$W = - 1 \times 8.314 \times 298 \times 2$

$W = - 4955.144 J$ = $- 4.955 k J$

Hence,option B is correct.

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