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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

For a hypothetical gaseous reaction at 300 K : X (g) arrow Y (g) +...

For a hypothetical gaseous reaction at 300 K300\text{ K}:
X (g)→Y (g)+2Z (g)X\text{ (g)} \rightarrow Y\text{ (g)} + 2Z\text{ (g)}
The standard internal energy change (ΔU⊖\Delta U^{\ominus}) is 15.02 kJ mol−115.02\text{ kJ mol}^{-1} and the standard entropy change (ΔS⊖\Delta S^{\ominus}) is 80 J K−1 mol−180\text{ J K}^{-1}\text{ mol}^{-1}.
Determine the standard Gibbs free energy change (ΔG⊖\Delta G^{\ominus}) and predict the spontaneity of the process at this temperature.
(Assume R=8.3 J K−1 mol−1R = 8.3\text{ J K}^{-1}\text{ mol}^{-1})

Options

  1. A−8.98 kJ mol−1-8.98\text{ kJ mol}^{-1}, spontaneous
  2. B+4.00 kJ mol−1+4.00\text{ kJ mol}^{-1}, non-spontaneous
  3. C−4.00 kJ mol−1-4.00\text{ kJ mol}^{-1}, spontaneous
  4. D+44.00 kJ mol−1+44.00\text{ kJ mol}^{-1}, non-spontaneous

Correct answer

C. −4.00 kJ mol−1-4.00\text{ kJ mol}^{-1}, spontaneous

Step-by-step solution

For the reaction X (g)→Y (g)+2Z (g)X\text{ (g)} \rightarrow Y\text{ (g)} + 2Z\text{ (g)}, the change in the number of moles of gaseous species is:

Δng=(1+2)−1=2\Delta n_g = (1 + 2) - 1 = 2

The standard enthalpy change is given by:

ΔH⊖=ΔU⊖+ΔngRT\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT

ΔH⊖=15.02+2×8.3×300×10−3\Delta H^{\ominus} = 15.02 + 2 \times 8.3 \times 300 \times 10^{-3}

ΔH⊖=15.02+4.98=20.00 kJ mol−1\Delta H^{\ominus} = 15.02 + 4.98 = 20.00\text{ kJ mol}^{-1}

The standard Gibbs free energy change is given by:

ΔG⊖=ΔH⊖−TΔS⊖\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

ΔG⊖=20.00−300×80×10−3\Delta G^{\ominus} = 20.00 - 300 \times 80 \times 10^{-3}

ΔG⊖=20.00−24.00=−4.00 kJ mol−1\Delta G^{\ominus} = 20.00 - 24.00 = -4.00\text{ kJ mol}^{-1}

Since ΔG⊖<0\Delta G^{\ominus} < 0, the reaction is spontaneous.

Answer: −4.00 kJ mol−1-4.00\text{ kJ mol}^{-1}, spontaneous
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