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Chapter 13: Problem 14

The rate of forward reaction is two times that of the backward reaction at a given temperature and identical concentration, \(K_{\text {equilimbrium }}\) is, (a) \(1.5\) (b) \(0.5\) (c) \(2.5\) (d) \(2.0\)

Short Answer

Expert verified
The equilibrium constant \(K\) is 2, therefore the correct answer is (d) 2.0.

Step by step solution

01

Understanding the Situation

In a reversible reaction, the rate of the forward reaction is related to the rate of the backward (or reverse) reaction. If it is known that the rate of the forward reaction is double that of the backward reaction, then, in essence, the ratio of the rate of forward reaction to the rate of backward reaction is 2:1.
02

Relating the Rates to the Equilibrium Constant

The equilibrium constant, denoted as K, is a measure of concentrations of products over reactants at equilibrium assuming that the reaction has the same molar coefficients for reactants and products. Since the rate of the forward reaction is double that of the backward reaction, the ratio of the forward reaction rate to the backward reaction rate is \(2/1 = 2\).
03

Pinpointing the Correct Answer

Since the ratio of the forward to backward reaction rates is 2 (which is also the value of the equilibrium constant \(K_{\text {equilibrium}}\)), the correct answer is \(K = 2\).

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Most popular questions from this chapter

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

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) \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)

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) \(K=10^{-2}\)

What will be the equilibrium constant at \(717 \mathrm{~K}\) for the reaction $$ 2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{~g}) $$ if its value for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \text { at } 717 \mathrm{~K} \text { is } 64 ? $$ (a) 8 (b) 64 (c) \(\frac{1}{64}\) (d) \(\frac{1}{8}\)
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