Pure gas 'A' was taken in a closed container at an initial pressure of 5 atm. The pressure in the container changed due to reactionA(g)⇌B(g)+2C(g)C(g)⇌B(g)+D(g)If the total pressure due to mixture of gas A, B. C and D was found to be 10 atm, in which pressure due to C was 3 atm. Find Kp for first equilibria.

Here, we have,A(g)⇌B(g)+2C(g)5−xx+y 2x−yC(g)⇌B(g)+D(g)2x−yx+y y2x−y=3 ...(i)5−x+x+y+2x+y=2 ...(ii)5+y+2x=10−y+2x=33−2y=7y=1,x=2Kp=PBPC2PA=3⋅323=9

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Pure gas 'A' was taken in a closed container at an initial pressure of 5 atm. The pressure in the container changed due to reaction
$\begin{array}{c} A (g) ⇌ B (g) + 2 C (g) \\ C (g) ⇌ B (g) + D (g) \end{array}$
If the total pressure due to mixture of gas A, B. C and D was found to be 10 atm, in which pressure due to C was 3 atm. Find Kp for first equilibria.

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answer is 9.

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Here, we have,

$\begin{array}{lc} A (g) ⇌ & B (g) + 2 C (g) \\ 5 - x & x + y 2 x - y \\ C (g) ⇌ & B (g) + D (g) \\ 2 x - y & x + y y \end{array}$

$\begin{array}{l} 2 x - y = 3 . . . (i) \\ 5 - x + x + y + 2 x + y = 2 . . . (i i) \\ 5 + y + 2 x = 10 \\ - y + 2 x = 3 \\ 3 - 2 y = 7 \end{array}$

$\begin{array}{l} y = 1, x = 2 \\ K_{p} = \frac{P_{B} P_{C}^{2}}{P_{A}} = \frac{3 \cdot 3^{2}}{3} = 9 \end{array}$

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