Question

At \(298\ K\), the equilibrium constant is \(2 × 10^{15}\) for the reaction:
\(Cu(s)+2Ag+(Aq)⇌Cu^{2+}(aq)+2Ag(s)\)
The equilibrium constant for the reaction:
\(\frac 12Cu^{2+}(aq)+Ag(s)+(Aq)⇌\frac 12Cu(s)+Ag^+(aq)\) is \(x \times10^{-8}\).
The value of \(x\) is ______ . (Nearest integer)

Answer 1

Correct Answer: 2

To solve the problem, we need to determine the equilibrium constant for the second reaction given the equilibrium constant of the first reaction.
The first reaction is:
\(Cu(s)+2Ag^+(aq)⇌Cu^{2+}(aq)+2Ag(s)\)
The equilibrium constant \(K_1\) for this reaction is \(2 \times 10^{15}\).

The second reaction is:
\(\frac{1}{2}Cu^{2+}(aq)+Ag(s)⇌\frac{1}{2}Cu(s)+Ag^+(aq)\)
Let the equilibrium constant for this reaction be \(K_2 = x \times 10^{-8}\).

By examining the reactions, we see that the second reaction is the reverse of half the first reaction. Thus, \(K_2\) is related to \(K_1\) by the equation:

\(K_2 = \left(K_1\right)^{-\frac{1}{2}}\)
Calculate \((K_1)^{-\frac{1}{2}}\):
\((2 \times 10^{15})^{-\frac{1}{2}} = \left(2^{-\frac{1}{2}}\right) \times \left(10^{15}\right)^{-\frac{1}{2}}\)
= \(\frac{1}{\sqrt{2}}\times 10^{-7.5}\)

Converting to the format \(x \times 10^{-8}\):
\(K_2 = \frac{1}{\sqrt{2}} \times 10^{-7.5} = \frac{1}{\sqrt{2}} \times \sqrt{10} \times 10^{-8}\)
\(\sqrt{10}\approx3.16\), hence:
\(\frac{1}{\sqrt{2}}\times3.16\approx2.23\)
Therefore, \(x\approx2.23\) can be rounded to the nearest integer:
\(x = 2\).

The value of \(x\) falls within the given range [2,2].