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Question 800010
JEE Chemistry · Hard

Chemical Thermodynamics and Thermochemistry: JEE Main MCQ

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JEE MainHardQuestion 800010

Question

For a reaction, ΔH = −890 kJ/mol and Δn₍g₎ = −1 at 25°C. Using ΔH = ΔU + Δn₍g₎RT (R = 8.314 J/mol K), ΔU is:
  1. A
    −887.5 kJ/mol
    Correct
  2. B
    −890 kJ/mol
  3. C
    −892.5 kJ/mol
  4. D
    −880 kJ/mol

Correct answer

−887.5 kJ/mol

Explanation

ΔU=ΔH−ΔngRT=−890−(−1)(8.314×10−3)(298)≈<strong>−887.5 kJ/mol</strong>ΔU=ΔH-Δn_gRT=-890-(-1)(8.314×10^{-3})(298)≈<strong>-887.5\text{ kJ/mol}</strong>.

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