2 mol of zinc is dissolved in HClat 25∘C. The work done in an open v

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Question

2 mol of zinc is dissolved in

H C l

at

25^{\circ} C

. The work done in an open vessel is:

Options:

A . −2.447 kJ

B . −4.955 kJ

C . R2

D . 3R

Answer: Option B
:
B
The following reaction takes place:

2 Z n + 4 H C l \to 2 Z n C l_{2} + 2 H_{2}

It is very impiortant to understand that this is an irreversible process since this reaction happens pretty quickly and hydrogengas escapes the system.
When

H_{2}

gas is liberated,it pushes back its surrounding and thus does some work against it.
Since it is an irreversible process, we can use the following equation:

W = P_{e x t} Δ V = - P_{e x t} (V_{f} - V_{x})

V_{i} = 0

thus

Δ V = V_{f}

V_{f} = \frac{n R T}{P_{e x t}}

W = - P e x t \times \frac{n R T}{P_{e x t}} = - n R T

Given that

n = 2, R = 8.314 J / K m o l

T = 25 + 273 = 298 K

W = - 2 \times 8.314 \times 298 = 4955.144 J = 4.955 k J

Tip: You can use approximations for

R = 25 / 3

and

T = 300

for quicker calculations!

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