Question

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$

Value of $K_c$ for the formation of ammonia according to the above equation is :

• A. $K_c=\frac{\left[N_2\right]\left[H_2\right]}{\left[N{H}_3\right]}$
• B. $K_c=\frac{\left[N{H}_3\right]}{\left[N_2\right]\left[H_2\right]}$
• C. $K_c=\frac{\left[N{]}_2\right[H{]}^6}{[N{H}_3{]}^2}$
• D. $K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$
• E. $K_c=\frac{\left[N_2\right][H_2{]}^3}{[N{H}_3{]}^2}$

Answer 1

Answer: (D)

$K_c=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$

If $aA+bB\rightleftarrows cC+dD$

then $Kc=\frac{{\left[C\right]}^c[D{]}^d}{{\left[A\right]}^a{\left[B\right]}^b}$

where $Kc$ = equilibrium constant.

$A,B$ = reactants

$C,D$ = products

$a,b,c,d$ = indices are the numbers in front of each substances in chemical equation.

In reaction,

$N_2\left(g\right)+3H_2\left(g\right)\rightleftarrows 2{NH}_3\left(g\right)$

$Kc=\frac{{\left[{NH}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$