1. Tardigrade
  2. Question
  3. Chemistry
  4. The value of equilibrium. constant of a reaction changes with change of temperature and the change is given by van’t Hoff equation, ( d ln K p / dT )=(Δ H °/ dT )=(Δ H 0/ RT 2) where enthalpy change, Δ H° is taken as constant in the small temperature range If for reaction, A(g)+3B(g) leftharpoons 2C(g), a plot of In Keq versus for 1/T a reaction is shown, then which of the following condition will be favourable for formation of product C ? <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/chemistry/2b0e1c2405b60e85c905325800040ab6-.png />

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Q. The value of equilibrium. constant of a reaction changes with change of temperature and the change is given by van’t Hoff equation, $\frac{ d \ln K _{ p }}{ dT }=\frac{\Delta \,H ^{\circ}}{ dT }=\frac{\Delta \,H ^{0}}{ RT ^{2}}$ where enthalpy change, $\Delta\,H^{\circ}$ is taken as constant in the small temperature range
If for reaction, $A_{(g)}+3B_{(g)} \rightleftharpoons 2C_{(g)}$, a plot of In $K_{eq}$ versus for $1/T$ a reaction is shown, then which of the following condition will be favourable for formation of product $C$ ?
image

Equilibrium

Solution:

Since slope is negative hence reaction is endothermic. So high temperature favours forward reaction similarly high pressure favours forward reaction