Calculate the work done in the following reaction at 300\ K , A(g) + B(g) = 2C(g) [R = 8.314\ JK^ -1 mol^ -1 ]

Correct answer: C. 00\ kJ The given reaction is A(g) + B(g) arrow 2C(g) . The change in the number of moles of gaseous species is given by Δ n_g = n_p - n_r . Δ n_g = 2 - (1 + 1) = 0 The work done in a chemical reaction at constant temperature and pressure is W = -Δ n_g RT . Substituting the value of Δ n_g , we get W = -0 × 8.314 × 300 = 0\ kJ . Answer: 00\ kJ

MHT CET 202624 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

MHT CET 2026 · 24 April · Evening Shift · Chemistry · Thermodynamics (C)

Calculate the work done in the following reaction at 300\ K , A(g)...

Calculate the work done in the following reaction at

300\ \text{K}

\text{A}(g) + \text{B}(g) = 2\text{C}(g)

[R = 8.314\ \text{JK}^{-1}\text{mol}^{-1}]

Options

A $2.494\ \text{kJ}$
B $4.988\ \text{kJ}$
C $00\ \text{kJ}$
D $-2.494\ \text{kJ}$

Correct answer

C. $00\ \text{kJ}$

Step-by-step solution

The given reaction is

\text{A}(g) + \text{B}(g) \rightarrow 2\text{C}(g)

.

The change in the number of moles of gaseous species is given by

\Delta n_g = n_p - n_r

\Delta n_g = 2 - (1 + 1) = 0

The work done in a chemical reaction at constant temperature and pressure is

W = -\Delta n_g RT

.

Substituting the value of

\Delta n_g

, we get

W = -0 \times 8.314 \times 300 = 0\ \text{kJ}

.

Answer:

00\ \text{kJ}

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