Answer
$2A(s)$ ⇌ $3D(g)$ , $K$ = $4.744\times10^{-4}$
Work Step by Step
$A(s)$(⇌$\frac{1}{2}$$B(g)$ + $C(g)$ $K_1$=0.0334
$3D(g)$⇌$B(g)$ + $2C(g)$ $K_2$ =2.35
1. In the final equation, the $D(g)$ molecule is on the products side. Thus, we should invert the second equation. The new equilibrium constant is equal to the multiplicative inverse of the old one.
$K_3$ = $\frac{1}{2.35}$ =0.4255
$A(s)$(⇌$\frac{1}{2}$$B(g)$ + $C(g)$ $K_1$=0.0334
$B(g)$ + $2C(g)$⇌ $3D(g)$ $K_3$ =0.4255
2. In the final equation the coefficient that goes with $A(s)$ is 2, so we should multiply all the first equation by 2. The new equilibrium constant is equal to 0.0334 to the power o 2.
$K_4$=$(0.0334)^{2}$=$1.115\times10^{-3}$
$2A(s)$⇌$B(g)$ + $2C(g)$ , $K_4$=$(0.0334)^{2}$=$1.115\times10^{-3}$
$B(g)$ + $2C(g)$⇌ $3D(g)$ $K_3$ =$0.4255$
3. Now, we can add the equations. The final equilibrium constant is equal to the multiplication of $1.115\times10^{-3}$ and $0.4255$
$K$=$1.115\times10^{-3}$$\times$$0.4255$
$K$ = $4.744\times10^{-4}$
$2A(s)$ + $B(g)$ + $2C(g)$⇌$B(g)$ + $2C(g)$ + $3D(g)$ , $K$ = $4.744\times10^{-4}$
$2A(s)$ ⇌ $3D(g)$ , $K$ = $4.744\times10^{-4}$
