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Exercises · 5.19

Q.For the reaction 2A(g)+B(g)→2D(g)2A(g) + B(g) \rightarrow 2D(g), ΔU=−10.5\Delta U = -10.5 kJ and ΔS=−44.1\Delta S = -44.1 JK−1^{-1}. Calculate ΔG\Delta G for the reaction, and predict whether the reaction may occur spontaneously.

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Using ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, we first find ΔH\Delta H from ΔU\Delta U via ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT, then compute ΔG\Delta G at 298 K. The result is positive, so the reaction is non-spontaneous at this temperature.

The key to solving this lies in connecting two thermodynamic quantities: internal energy change (ΔU\Delta U) and enthalpy change (ΔH\Delta H), and then using Gibbs free energy to judge spontaneity. You're given ΔU\Delta U and ΔS\Delta S, but the Gibbs equation uses ΔH\Delta H, not ΔU\Delta U. So the first step is always to convert.

Why? Because ΔU\Delta U is measured at constant volume, while most reactions (including this one) occur at constant pressure (open container). The enthalpy change ΔH\Delta H accounts for the pressure-volume work done by or on the system. The relation is:

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

where Δng\Delta n_g is the change in moles of gas.

Let's work through it step by step.

  1. Find Δng\Delta n_g

    For the reaction 2A(g)+B(g)→2D(g)2A(g) + B(g) \rightarrow 2D(g):

    Moles of gaseous products = 2

    Moles of gaseous reactants = 2 + 1 = 3

    So Δng=2−3=−1\Delta n_g = 2 - 3 = -1.

  2. Calculate ΔH\Delta H

    Given ΔU=−10.5\Delta U = -10.5 kJ = −10500-10500 J (we'll work in J for consistency with ΔS\Delta S in J/K).

    Use R=8.314R = 8.314 J mol−1^{-1} K−1^{-1} and assume standard temperature T=298T = 298 K (since not specified, this is the default for such problems).

ΔH=ΔU+ΔngRT=−10500+(−1)(8.314)(298)\Delta H = \Delta U + \Delta n_g RT = -10500 + (-1)(8.314)(298)

Compute 8.314×298=2477.5728.314 \times 298 = 2477.572 J.

So ΔH=−10500−2477.572=−12977.572\Delta H = -10500 - 2477.572 = -12977.572 J ≈−12.98\approx -12.98 kJ.

Note

The negative ΔH\Delta H tells us the reaction is exothermic — it releases heat. But that alone doesn't guarantee spontaneity; entropy also matters.

  1. Apply the Gibbs free energy equation

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Given ΔS=−44.1\Delta S = -44.1 J K−1^{-1}.

At T=298T = 298 K:

ΔG=(−12977.572)−(298)(−44.1)\Delta G = (-12977.572) - (298)(-44.1)

Compute 298×(−44.1)=−13141.8298 \times (-44.1) = -13141.8 J.

So:

ΔG=−12977.572−(−13141.8)=−12977.572+13141.8=164.228 J\Delta G = -12977.572 - (-13141.8) = -12977.572 + 13141.8 = 164.228 \text{ J}

That's about 0.1640.164 kJ. …

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