First slide

Work

Question

The work done when 65.38g of zinc dissolved completely in HCl in an open beaker at 300k is

Easy

A

– 2494.2J
B

– 249.4J
C

1J
D

– 24.94J

Solution

Zn + 2HCl → ZnCl₂+H₂ ↑
No. of moles of Zn = No. of mole of H₂ liberated $= \frac{65 .38}{65.38} = 1 mole$
                              $V_{H_{2}} = \frac{n R T}{P} = \frac{1 \times 0 .0821 \times 300}{1 atm} = 24 .63 lit$
w = - PΔV = - 1 atm [24.63-0]
   = -24.63 x 101.3 J
   = - 2495J

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