Answer
$2A(s) ⇌ 3D(g) , K = 4.744×10^{-4}$
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 of 2.
$K_4=(0.0334)^{2}=1.115×10^{-3}$
$2A(s)⇌B(g) + 2C(g) , K_4=(0.0334)^{2}=1.115×10^{-3}$
$B(g) + 2C(g)⇌ 3D(g), K_3 =0.4255$
3. Now, we can add the equations. The final equilibrium constant is equal to the multiplication of $1.115×10^{-3}$ and $0.4255$
$K=1.115×10^{-3}×0.4255$
$K = 4.744×10^{-4}$
$2A(s) + B(g) + 2C(g)⇌B(g) + 2C(g) + 3D(g)$ , $K = 4.744×10^{-4}$
$2A(s) ⇌ 3D(g)$ , $K = 4.744×10^{-4}$
