Question

According to the following reaction:

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

$4$ moles of $A$ reacts with $4$ moles of $B$ to form $2$ moles of $C$ at equilibrium. What is the value of the equilibrium constant?

Answer 1

Explanation:

Equation: $A_{\left(g\right)}+B_{\left(g\right)}\to C_{\left(g\right)}+D_{\left(g\right)}$

$4$ $4$ $0$ $0$ At time $t=0$

$4-x$ $4-x$ $x$ $x$ At equilibrium

If $2$ moles of C are found at equilibrium then $x=2$

Remaining moles of $A=4-x=4-2=2$

Remaining moles of $B=4-x=4-2=2$

Moles of D formed $=2$

The equilibrium constant $K_C=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

$K_c=\frac{2\times 2}{2\times 2}=1$

Hence the correct answer is $K_C=1$