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Chapter 13: Problem 9

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) \(0.02\)

Short Answer

Expert verified
The equilibrium constant (K) for the dissociation of HI will be \(0.02\). Therefore, option (d) is the correct answer.

Step by step solution

01

Understanding the Synthesis Reaction

The synthesis of hydrogen iodide (HI) is given by the reaction: \(H_{2}(g) + I_{2}(g) \rightarrow 2HI(g)\). It's also given that the equilibrium constant (\(K\)) for this reaction is 50 at 490 °C.
02

Understanding the Dissociation Reaction

The dissociation of HI is simply the reverse of the synthesis reaction: \(2HI(g) \rightarrow H_{2}(g) + I_{2}(g)\). The equilibrium constant for this reverse reaction is what needs to be found.
03

Find the Equilibrium Constant for the Dissociation Reaction

The equilibrium constant of a reverse reaction is equal to the reciprocal of the equilibrium constant of the original reaction. So, for the dissociation reaction, the equilibrium constant (\(K'\)) can be found by taking the reciprocal of the equilibrium constant of the synthesis reaction. Mathematically, \(K' = 1/K = 1/50 = 0.02\)

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Most popular questions from this chapter

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) \(0.026\)

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 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) \(\frac{1}{8}\)

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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