Respuesta :
The question is incomplete, here is the complete question:
Consider the reaction [tex]4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)[/tex]
Initially, [tex][NH_3(g)]=[O_2(g)][/tex] = 3.60 M; at equilibrium [tex][N_2O_4(g)]=0.60M[/tex] . Calculate the equilibrium concentration for [tex]NH_3[/tex]
Answer: The equilibrium concentration of ammonia is 2.8 M
Explanation:
We are given:
Initial concentration of [tex][NH_3(g)][/tex] = 3.60 M
Initial concentration of [tex][O_2(g)][/tex] = 3.60 M
For the given chemical equation:
[tex]4NH_3(g)+7O_2(g)\rightarrow 2N_2O_4(g)+6H_2O(g)[/tex]
Initial: 3.60 3.60
At eqllm: 3.60-4x 3.60-7x 2x 6x
We are given:
Equilibrium concentration of [tex][N_2O_4(g)][/tex] = 0.60 M
Evaluating the value of 'x'
[tex]\Rightarrow 2x=0.60\\\\\Rightarrow x=\frac{0.60}{2}=0.2[/tex]
So, equilibrium concentration of [tex]NH_3=(3.60-4x)=(3.60-(4\times 0.2))=2.8M[/tex]
Hence, the equilibrium concentration of ammonia is 2.8 M
