JEE MAIN - Chemistry (2022 - 26th July Morning Shift - No. 20)
At $$298 \mathrm{~K}$$, the equilibrium constant is $$2 \times 10^{15}$$ for the reaction :
$$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$$
The equilibrium constant for the reaction
$$ \frac{1}{2} \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) $$
is $$x \times 10^{-8}$$. The value of $$x$$ is _____________. (Nearest Integer)
Answer
2
Explanation
$\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
$k=2 \times 10^{15}$
$\frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
$\mathrm{K}^{\prime}=\frac{1}{(\mathrm{~K})^{1 / 2}}=\frac{1}{\left(2 \times 10^{15}\right)^{1 / 2}}$
$=2.23 \times 10^{-8}$
$x \simeq 2$
$k=2 \times 10^{15}$
$\frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
$\mathrm{K}^{\prime}=\frac{1}{(\mathrm{~K})^{1 / 2}}=\frac{1}{\left(2 \times 10^{15}\right)^{1 / 2}}$
$=2.23 \times 10^{-8}$
$x \simeq 2$
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