Respuesta :
Answer: The value of [tex]K_c[/tex] for [tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex] reaction is [tex]5.13\times 10^2[/tex]
Explanation:
We are given:
Initial moles of nitrogen gas = 1.30 moles
Initial moles of hydrogen gas = 1.65 moles
Equilibrium moles of ammonia = 0.100 moles
Volume of the container = 1.00 L
For the given chemical equation:
[tex]N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
Initial: 1.30 1.65
At eqllm: 1.30-x 1.65-3x 2x
Evaluating the value of 'x'
[tex]\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}[/tex]
Equilibrium moles of nitrogen gas = [tex](1.30-x)=(1.30-0.05)=1.25mol[/tex]
Equilibrium moles of hydrogen gas = [tex](1.65-x)=(1.65-0.05)=1.60mol[/tex]
Putting values in above expression, we get:
[tex]K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}[/tex]
Calculating the [tex]K_c'[/tex] for the given chemical equation:
[tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex]
[tex]K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2[/tex]
Hence, the value of [tex]K_c[/tex] for [tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex] reaction is [tex]5.13\times 10^2[/tex]
Answer:
[N2] = 1.25 M
[H2] = 1.50 M
[NH3] = 0.100 M
Kc = 421.875
Explanation:
Step 1: Data given
Volume = 1.00L
Temperature = 727 °C
Number of moles N2 = 1.30 moles
Number of moles H2 = 1.65 moles
At equilibrium, 0.100 mol of NH3 is present
Step 2: The balanced equation
2 NH3(g) ⇌ N2(g) + 3 H2(g)
N2(g) + 3H2(g) ⇌2NH3(g)
Step 3: The initial moles
N2 = 1.30 moles
H2 = 1.65 moles
NH3 = 0 moles
Step 4: Calculate moles at the equilibrium
For 2 moles NH3 we need 1 mol N2 and 3 moles H2
N2 = 1.30 - x
H2 = 1.65 - 3x
NH3 = 2x = 0.100 moles
x = 0.100 /2 = 0.050 moles
N2 = 1.30 - 0.050 = 1.25 moles
H2 = 1.65 - 3*0.050 = 1.50 moles
NH3 = 2x = 0.100 moles
Step 5: Calculate Kc
2NH3(g) ⇌ N2(g) + 3H2(g)
Kc = [N2][H2]³/[NH3]²
Kc = (1.25 * 1.50³)/0.100²
Kc = 421.875
