Chemical equilibrium

Question

37. Consider the following equilibria: \[ \begin{array}{ll} A_{(s)} \rightleftharpoons B_{(g)}+C_{(g)} & \left(K_{p}\right)_{1}=x \\ B_{(g)} \rightleftharpoons D_{(g)} & \left(K_{p}\right)_{2}=2 \end{array} \] Initially only \( A _{( s )} \) was present and final total pressure at equilibrium is \( 20 atm \). Equilibrium constant \( (x) \) is found to be A) 100 B) 33.33 C) 50 D) 24

Answer 1

A_(s)-->B_(g) + C_(g)

B_(g)-->D_(g)

​Adding these​​, we get:

​A_(s)-->D_(g)+C_(g)

Taking P as initial Pressure of A and y as the amount dissociated,

A_(s)⇔D_(g)+C_(g)

P-y         y         y

Total pressure=20 (neglecting pressure of solid)

y+y=20

y= 10   

K_p= K_p1×K_p2

_Kp=2x

_y²=2x

_100=2x

_⇒x=50

_{Hence, 'C'}