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Thermodynamics question

2018 · Shift 0 · Q22
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Thermodynamics question

2018 · Shift 0 · Q22

JEE MainChemistryThermodynamicsMCQ+4 / −1
Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction? JEE Main 2018 (Offline) Chemistry - Thermodynamics Question 180 English
  1. A
    A and D
  2. B
    A and B
  3. C
    B and C
  4. D
    C and D
View written solutionFree

Correct answer: B

  1. For the temperature dependence of equilibrium constant, we use the van’t Hoff equation:
d(ln⁡K)dT=ΔHRT2\frac{d(\ln K)}{dT} = \frac{\Delta H}{RT^2}dTd(lnK)​=RT2ΔH​

For an exothermic reaction,

ΔH<0\Delta H < 0ΔH<0

So,

d(ln⁡K)dT<0\frac{d(\ln K)}{dT} < 0dTd(lnK)​<0

This means:

  • As temperature TTT increases, ln⁡K\ln KlnK decreases.
  • Hence, KKK also decreases with increase in TTT.
  1. Therefore, the correct graphical trends for an exothermic reaction must show:
  • KKK decreasing with TTT, or
  • ln⁡K\ln KlnK decreasing with 1/T1/T1/T according to the appropriate van’t Hoff form.

Using the integrated van’t Hoff equation:

ln⁡K=−ΔHR⋅1T+C\ln K = -\frac{\Delta H}{R}\cdot \frac{1}{T} + ClnK=−RΔH​⋅T1​+C

Since ΔH<0\Delta H < 0ΔH<0, we get

−ΔH>0-\Delta H > 0−ΔH>0

So the slope of the graph of ln⁡K\ln KlnK vs 1/T1/T1/T is positive.

Thus, the correct lines are:

  • one showing KKK decreases with TTT
  • one showing ln⁡K\ln KlnK increases linearly with 1/T1/T1/T

These correspond to A and B.

  1. Hence the correct option is:
B\boxed{\text{B}}B​
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