# 10 how does the value of the equilibrium constant depend on temperature? Ideas Contents

Below is information and knowledge on the topic how does the value of the equilibrium constant depend on temperature? gather and compiled by the show.vn team. Along with other related topics like: What happens to equilibrium when temperature is increased, What happens to equilibrium when temperature is increased endothermic, What happens to equilibrium when temperature is increased exothermic, What happens to equilibrium when temperature is decreased, Does concentration affect equilibrium constant, How does concentration affect equilibrium, Temperature dependence of equilibrium constant formula, Does pressure affect equilibrium constant. erature Dependence of Equilibrium Constants – the van ’t Hoff Equation

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• The value of $$K_p$$ is independent of pressure, although the composition of a system at equilibrium may be very much dependent on pressure. Temperature dependence is another matter. Because the value of $$\Delta G_{rxm}^o$$ is dependent on temperature, the value of $$K_p$$ is as well. The form of the temperature dependence can be taken from the definition of the Gibbs function. At constant temperature and pressure

$\dfrac{\Delta G^o_{T_2}}{T_2} – \dfrac{\Delta G^o_{T_1}}{T_1} = \Delta H^o \left(\dfrac{1}{T_2} – \dfrac{1}{T_1} \right)$

Substituting

$\Delta G^o = -RT \ln K$

For the two values of $$\Delta G_{}^o$$ and using the appropriate temperatures, yields

$\dfrac{-R{T_2} \ln K_2}{T_2} – \dfrac{-R{T_1} \ln K_1}{T_1} = \Delta H^o \left(\dfrac{1}{T_2} – \dfrac{1}{T_1} \right)$

And simplifying the expression so that only terms involving $$K$$ are on the left and all other terms are on the right results in the van ’t Hoff equation, which describes the temperature dependence of the equilibrium constant.

$\ln \left(\dfrac{\ K_2}{\ K_1}\right) = – \dfrac{\Delta H^o}{R} \left(\dfrac{1}{T_2} – \dfrac{1}{T_1} \right) \label{vH}$

Because of the assumptions made in the derivation of the Gibbs-Helmholtz equation, this relationship only holds if $$\Delta H^o$$ is independent of temperature over the range being considered. This expression also suggests that a plot of $$\ln(K)$$ as a function of $$1/T$$ should produce a straight line with a slope equal to $$–\Delta H^o/R$$. Such a plot is known as a van ’t Hoff plot, and can be used to determine the reaction enthalpy.

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Example $$\PageIndex{1}$$

A certain reaction has a value of $$K_p = 0.0260$$ at 25 °C and $$\Delta H_{rxm}^o = 32.4 \,kJ/mol$$. Calculate the value of $$K_p$$ at 37 °C.

Solution

This is a job for the van ’t Hoff equation!

• T1 = 298 K
• T2 = 310 K
• $$\Delta H_{rxm}^o = 32.4 \,kJ/mol$$
• K1 = 0.0260
• K2 = ?

So Equation \ref{vH} becomes

\begin{align*} \ln \left( \dfrac{\ K_2}{0.0260} \right) &= – \dfrac{32400 \,J/mol}{8.314 \,K/(mol \,K)} \left(\dfrac{1}{310\, K} – \dfrac{1}{298 \,K} \right) \\[4pt] K_2 &= 0.0431 \end{align*}

Note: the value of $$K_2$$ increased with increasing temperature, which is what is expected for an endothermic reaction. An increase in temperature should result in an increase of product formation in the equilibrium mixture. But unlike a change in pressure, a change in temperature actually leads to a change in the value of the equilibrium constant!

Example $$\PageIndex{2}$$

Given the following average bond enthalpies for $$\ce{P-Cl}$$ and $$\ce{Cl-Cl}$$ bonds, predict whether or not an increase in temperature will lead to a larger or smaller degree of dissociation for the reaction

$\ce{PCl_5 \rightleftharpoons PCl_3 + Cl_2} \nonumber$

X-Y D(X-Y) (kJ/mol)
P-Cl 326
Cl-Cl 240

Solution

The estimated reaction enthalpy is given by the total energy expended breaking bonds minus the energy recovered by the formation of bonds. Since this reaction involves breaking two P-Cl bonds (costing 652 kJ/mol) and the formation of one Cl-Cl bond (recovering 240 kJ/mol), it is clear that the reaction is endothermic (by approximately 412 kJ/mol). As such, an increase in temperature should increase the value of the equilibrium constant, causing the degree of dissociation to be increased at the higher temperature.

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## Frequently Asked Questions About how does the value of the equilibrium constant depend on temperature?

If you have questions that need to be answered about the topic how does the value of the equilibrium constant depend on temperature?, then this section may help you solve it.

### Is K equilibrium temperature-dependent?

The magnitude of K decreases with increasing temperature for exothermic reactions (?H° 0), whereas for endothermic reactions (?H° > 0), the magnitude of K increases with increasing temperature, assuming that?H° and?S° are temperature independent.

### How is K equilibrium impacted by temperature?

The equilibrium will shift in favor of reactant formation and a decrease in Keq as a result of an increase in temperature, which is why the answer is “Keq will decrease.”

### Why does temperature cause the K value to rise?

The Arrhenius Equation states that as T increases, the value of the exponential part of the equation becomes less negative, increasing the value of k. This means that increasing the temperature of a reaction generally speeds up the process (increases the rate) because the rate constant increases.

### What occurs to constant K as the temperature rises?

The reaction is exothermic because the K value decreases as the temperature rises.