a box contains ammonia, hydrogen, and nitrogen gases at equilibrium at 1,000 k. analysis gives the following…

a box contains ammonia, hydrogen, and nitrogen gases at equilibrium at 1,000 k. analysis gives the following concentrations at equilibrium: h₂ = 2.0 m, n₂ = 3.0 m, nh₃ = 2.0 m. calculate the value of the equilibrium constant for: 2 nh₃(g) ⇌ n₂(g) + 3 h₂(g)\n0.17\n0.33\n0.77\n2.25\n3.00\n4.33\n5.25\n6.00

a box contains ammonia, hydrogen, and nitrogen gases at equilibrium at 1,000 k. analysis gives the following concentrations at equilibrium: h₂ = 2.0 m, n₂ = 3.0 m, nh₃ = 2.0 m. calculate the value of the equilibrium constant for: 2 nh₃(g) ⇌ n₂(g) + 3 h₂(g)\n0.17\n0.33\n0.77\n2.25\n3.00\n4.33\n5.25\n6.00

Answer

Explanation:

Step1: Write equilibrium - constant expression

The equilibrium - constant expression ($K_c$) for the reaction $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ is $K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}$.

Step2: Substitute the given concentrations

We are given that $[H_2] = 2.0\ M$, $[N_2]=3.0\ M$, and $[NH_3]=2.0\ M$. Substitute these values into the equilibrium - constant expression: $K_c=\frac{(3.0)\times(2.0)^3}{(2.0)^2}$.

Step3: Simplify the expression

First, calculate $(2.0)^3 = 8.0$ and $(2.0)^2 = 4.0$. Then the expression becomes $K_c=\frac{3.0\times8.0}{4.0}$. $3.0\times8.0 = 24.0$, and $\frac{24.0}{4.0}=6.0$.

Answer:

$6.00$