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Chapter 13: Problem 7

With increase in temperature, equilibrium constant of a reaction (a) always decreases (b) always increases (c) may increase or decrease depending upon whether \(n_{p}n_{r}\).

Short Answer

Expert verified
The equilibrium constant may increase or decrease with increasing temperature depending on whether the reaction is endothermic (\(n_{p}>n_{r}\)) or exothermic (\(n_{p}<n_{r}\)), hence the correct answer is (c).

Step by step solution

01

Interpreting the effect of temperature on equilibrium constant

According to the Van't Hoff equation, the equilibrium constant of a reaction is directly proportional to the changes in temperature for a reaction. When the temperature increases, if the reaction is exothermic (i.e., releases heat), the equilibrium shifts to the left (reactant side), meaning the equilibrium constant decreases, and hence option (a) is incorrect. Conversely, if the reaction is endothermic (i.e., absorbs heat), the equilibrium shifts to the right (product side), which leads to the increase of the equilibrium constant, and hence option (b) is also incorrect.
02

Connecting changes in moles with temperature on equilibrium constant

We need to connect this idea with whether \(n_{p}n_{r}\). In a chemical reaction, if there are more moles on the product side (\(n_{p} > n_{r}\)), the reaction is endothermic. Thereby, when the temperature increases, the equilibrium constant increases. On the contrary, when there are more moles on the reactants side (\(n_{p}<n_{r}\)), the reaction is exothermic. Thus, an increase in temperature results in a decrease in the equilibrium constant. This connects to the changes of the equilibrium constant with the number of moles of reactants and products under different temperature conditions.
03

Conclusion of the exercise

Hence, the equilibrium constant may increase or decrease with increasing temperature depending on whether the reaction is endothermic (\(n_{p}>n_{r}\)) or exothermic (\(n_{p}<n_{r}\)). This satisfies option (c) in the exercise.

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Most popular questions from this chapter

In which of the following reaction is \(K_{p}\) less than \(K_{c} ?\) (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (b) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) (d) \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)

In a reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PC}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) moles/litre respectively. If the value of \(K_{c}\) is \(0.5\), what is the concentration of \(\mathrm{C} 1_{2}\) in mole/litre? (a) \(1.0\) (b) \(1.5\) (c) \(0.5\) (d) \(2.0\)

At \(490^{\circ} \mathrm{C}\), the equilibrium constant for the synthesis of \(\mathrm{HI}\) is 50 . The value of \(K\) for the dissociation of HI will be (a) \(2.0\) (b) \(20.0\) (c) \(0.002\) (d) \(0.02\)

In which of the following cases does the reaction go farthest to completion? (a) \(K=10\) (b) \(K=1\) (c) \(K=10^{3}\) (d) \(K=10^{-2}\)

In a reaction $$ A+2 B \rightleftharpoons 2 C $$ \(2.0\) moles of ' \(A\) ', \(3.0\) moles of ' \(B\) ' and \(2.0\) moles of ' \(C\) ' are placed in a \(2.0\) L flask and the equilibrium concentration of ' \(C\) ' is \(0.5\) mole/L. The equilibrium constant \((K)\) for the reaction is (a) \(0.147\) (b) \(0.073\) (c) \(0.05\) (d) \(0.026\)
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