Question

The equilibrium constant for the following reactions are given at \(25^{\circ} \mathrm{C}\)
$$
\begin{aligned}
&2 A \rightleftharpoons B+C, K_{1}=1.0 \\
&2 B \rightleftharpoons C+D, K_{2}=16 \\
&2 C+D \rightleftharpoons 2 P, K_{3}=25
\end{aligned}
$$
The equilibrium constant for the reaction \(\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}\) at \(25^{\circ} \mathrm{C}\) is
(A) \(\frac{1}{20}\)
(B) 20
(C) \(\frac{1}{42}\)
(D) 21

Answer 1

Ans: \((A)\)
 We can manipulate the given equations as follows
$$
\begin{array}{ll}
\frac{1}{2} \mathrm{~B}+\frac{1}{2} \mathrm{C} \rightleftharpoons \mathrm{A}, & \mathrm{K}_{1}^{\prime}=1 \\
\frac{1}{2} \mathrm{C}+\frac{1}{2} \mathrm{D} \rightleftharpoons \mathrm{B}, & \mathrm{K}_{2}^{\prime}=\frac{1}{\sqrt{16}} \\
\mathrm{P} \rightleftharpoons \mathrm{C}+\frac{1}{2} \mathrm{D}, & \mathrm{K}_{3}^{\prime}=\frac{1}{\sqrt{25}}
\end{array}
$$
$$
\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}, \quad \mathrm{~K}_{\text {final }}=1 \times \frac{1}{4} \times \frac{1}{5}=\frac{1}{20}
$$

Answer 2

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Answer 3

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