Question

Initially, [NH3(g)] = [O2(g)] = 3.60 M; at equilibrium [N2O4(g)] = 0.60 M. Calculate the equilibrium concentration for NH3.

Answer 1

The question is incomplete, here is the complete question:

Consider the reaction  $4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initially, $[NH_3(g)]=[O_2(g)]$ = 3.60 M; at equilibrium $[N_2O_4(g)]=0.60M$  . Calculate  the equilibrium concentration for $NH_3$

Answer: The equilibrium concentration of ammonia is 2.8 M

Explanation:

We are given:

Initial concentration of $[NH_3(g)]$ = 3.60 M

Initial concentration of $[O_2(g)]$ = 3.60 M

For the given chemical equation:

                     $4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)$

Initial:               3.60        3.60            

At eqllm:      3.60-4x     3.60-7x           2x             6x

We are given:

Equilibrium concentration of $[N_2O_4(g)]$ = 0.60 M

Evaluating the value of 'x'

$\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2$

So, equilibrium concentration of $NH_3=(3.60-4x)=(3.60-(4\times 0.2))=2.8M$

Hence, the equilibrium concentration of ammonia is 2.8 M