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Chapter 13: Problem 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) \(\frac{1}{8}\)

Short Answer

Expert verified
\(\frac{1}{64}\)

Step by step solution

01

Identify the relationship between the reactions

The first reaction given is the reverse of the second reaction. Thus, the equilibrium constant of the first reaction would be the reciprocal of the equilibrium constant for the second reaction.
02

Use the equilibrium constant for the second reaction

According to the problem, the equilibrium constant for the reaction \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)\) at \(717 K\) is given to be \(64\). We denote this as \(K_2 = 64\).
03

Solve for the equilibrium constant of the first reaction

As explained earlier, the equilibrium constant of the reverse reaction is the reciprocal of that of the original reaction. Hence, the equilibrium constant for the first reaction (denoted as \(K_1\)) is \(\frac{1}{K_2} = \frac{1}{64}\).

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

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 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) \(0.026\)

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

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) \(2.0\)
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