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Chemical Kinetics and Nuclear Chemistry question

2005 · Shift 0 · Q27
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Chemical Kinetics and Nuclear Chemistry question

2005 · Shift 0 · Q27

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
A schematic plot of lnKeqln{K_{eq}}lnKeq​ versus inverse of temperature for a reaction is shown below AIEEE 2005 Chemistry - Chemical Kinetics and Nuclear Chemistry Question 158 English The reaction must be
  1. A
    highly spontaneous at ordinary temperature
  2. B
    one with negligible enthalpy change
  3. C
    endothermic
  4. D
    exothermic
View written solutionFree

Correct answer: D

  1. For a reaction, the temperature dependence of equilibrium constant is given by the van’t Hoff equation:
ln⁡Keq=−ΔH∘R⋅1T+ΔS∘R\ln K_{eq} = -\frac{\Delta H^\circ}{R}\cdot \frac{1}{T} + \frac{\Delta S^\circ}{R}lnKeq​=−RΔH∘​⋅T1​+RΔS∘​
  1. This is of the form:
y=mx+cy = mx + cy=mx+c

with

  • y=ln⁡Keqy = \ln K_{eq}y=lnKeq​
  • x=1Tx = \frac{1}{T}x=T1​
  • slope m=−ΔH∘Rm = -\frac{\Delta H^\circ}{R}m=−RΔH∘​
  1. From the given schematic plot of ln⁡Keq\ln K_{eq}lnKeq​ versus 1/T1/T1/T, the line has positive slope.

So,

−ΔH∘R>0-\frac{\Delta H^\circ}{R} > 0−RΔH∘​>0

Since R>0R>0R>0,

ΔH∘<0\Delta H^\circ < 0ΔH∘<0

Thus, the reaction is exothermic.

  1. Now check the options:

  • A: highly spontaneous at ordinary temperature
    This cannot be concluded only from the slope. Spontaneity depends on ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln KΔG∘=−RTlnK, and exact value of KKK at ordinary temperature is needed.

  • B: one with negligible enthalpy change
    If enthalpy change were negligible, slope would be nearly zero. But the graph has non-zero slope.

  • C: endothermic
    Endothermic means ΔH∘>0\Delta H^\circ > 0ΔH∘>0, which would give negative slope. Incorrect.

  • D: exothermic
    Correct, since ΔH∘<0\Delta H^\circ < 0ΔH∘<0.

  1. Therefore, the correct answer is:
D: exothermic\boxed{\text{D: exothermic}}D: exothermic​
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