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

Step-by-step 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=(0.60)/(2)=0.2`

So, equilibrium concentration of
`NH_3=(3.60-4x)=(3.60-(4* 0.2))=2.8M`

Hence, the equilibrium concentration of ammonia is 2.8 M