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MHT CET 202626 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
26 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

MHT CET 2026 · 26 April · Morning Shift · Chemistry · Thermodynamics (C)

Calculate Δ H for a reaction, 2X_ (g) + Y_ (g) arrow 2Z_ (g) at 298...

Calculate ΔH\Delta H for a reaction, 2X(g)+Y(g)→2Z(g)2X_{(g)} + Y_{(g)} \rightarrow 2Z_{(g)} at 298298 K if ΔU=−10.5\Delta U = -10.5 kJ and R=8.314 J K−1 mol−1R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}.

Options

  1. A−14.45-14.45 kJ
  2. B−12.98-12.98 kJ
  3. C10.5010.50 kJ
  4. D11.0011.00 kJ

Correct answer

B. −12.98-12.98 kJ

Step-by-step solution

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

Δng=moles of gaseous products−moles of gaseous reactants\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}

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

The relationship between enthalpy change (\Delta H)andinternalenergychange(ΔU) and internal energy change (\Delta U) is given by:

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

Substituting the given values:

ΔH=−10.5 kJ+(−1)×(8.314×10−3 kJ K−1 mol−1)×298 K\Delta H = -10.5 \text{ kJ} + (-1) \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times 298 \text{ K}

ΔH=−10.5 kJ−2.477 kJ\Delta H = -10.5 \text{ kJ} - 2.477 \text{ kJ}

ΔH=−12.977 kJ≈−12.98 kJ\Delta H = -12.977 \text{ kJ} \approx -12.98 \text{ kJ}

Answer: −12.98-12.98 kJ
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