Question

Consider the reaction: P(s) + 3/2 Cl2(

Answer 1

Answer: The equilibrium constant for the total reaction is $K=\frac{K_a}{K_b}$

Explanation:

We are given two intermediate equations:

Equation 1:  $P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g);K_a$

The expression of $K_a$ for the above equation is:

$K_a=\frac{[PCl_5]}{[P][Cl_2]^{5/2}}$       ......(1)

Equation 2:  $PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g);K_b$

The expression of $K_b$ for the above equation is:

$K_b=\frac{[PCl_5]}{[PCl_3][Cl_2]}$       ......(2)

Now, dividing expression 1 by expression 2, we get:

$\frac{K_a}{K_b}=\left(\frac{\frac{[PCl_5]}{[P][Cl_2]^{5/2}}}{\frac{[PCl_5]}{[PCl_3][Cl_2]}}\right)\\\\\\\frac{K_a}{K_b}=\frac{[PCl_3]}{[P][Cl_2]^{3/2}}$

The above expression is the expression for the total equation, which is:

$P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g);K$

Hence, the equilibrium constant for the total reaction is $K=\frac{K_a}{K_b}$

Answer 2

K = Ka/Kb

Further explanation

Given:

• $\boxed{ \ P_{(s)} + \frac{5}{2} Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_a \ }$
• $\boxed{ \ PCl_3_{(g)} + Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_b \ }$

Question:

Write the equilibrium constant for this reaction:

$\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = ? \ }$

in terms of the equilibrium constants, Ka and Kb

The Process:

Let us solve the problem above.

$\boxed{ \ P_{(s)} + \frac{5}{2} Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_a \ }$ ... (Equilibrium-1)

$\boxed{ \ PCl_3_{(g)} + Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_b \ }$ ... (Equilibrium-2)

Consider the reaction: $\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = ? \ }$

Equilibrium-2 is reversed to match the target reaction (Hint: PCl₃). Then the two equilibrium reactions are added together.

$\boxed{ \ P_{(s)} + \frac{5}{2} Cl_2_{(g)} \rightleftharpoons PCl_5_{(g)} \ \ \ \ \ K_a \ }$ ... (Equilibrium-1)

$\boxed{ \ PCl_5_{(g)} \rightleftharpoons PCl_3_{(g)} + Cl_2_{(g)} \ \ \ \ \ \frac{1}{K_b} \ }$ ... (Equilibrium-2, after reversed)

---------------------------------------------------------------- (+)

$\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = K_a \times \frac{1}{K_b} \ }$

Notes:

• $\boxed{PCl_5}$ is eliminated.
• $\boxed{\frac{5}{2}Cl_2 \ subtracted \ by \ Cl_2 \ equal \ to \ \frac{3}{2}Cl_2}$.
• If the two equilibrium reactions are added together, the equilibrium constants are multiplied by each other.

Thus, the equilibrium constant for the target reaction in terms of the equilibrium constants, Ka and Kb, is

$\boxed{\boxed{ \ P_{(s)} + \frac{3}{2} Cl_2_{(g)} \rightleftharpoons PCl_3_{(g)} \ \ \ \ \ K = \frac{K_a}{K_b} \ }}$

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Keywords: consider the reaction, P, Cl2, PCl3, PCl5, Ka, Kb, K, write, the equilibrium constant, for this, in terms, reverse, multiply