Question

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol of N2 and 1.65 mol of H2 are added. At equilibrium, 0.100 mol of NH3 is present. Calculate the equilibrium concentrations of N2 and H2 and find Kc for the reaction: 2 NH3(g) ⇌ N2(g) + 3 H2(g)

Answer 1

[N2] = 1.25 M

[H2] = 1.50 M

[NH3] = 0.100 M

Kc = 421.875

Step-by-step 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

Answer 2

Answer: The value of
`K_c` for
`2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)` reaction is
`5.13* 10^2`

Step-by-step 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:

`N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)`

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

`\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol`

The expression of
`K_c` for above equation follows:

`K_c=([NH_3]^2)/([N_2]* [H_2]^3)`

Equilibrium moles of nitrogen gas =
`(1.30-x)=(1.30-0.05)=1.25mol`

Equilibrium moles of hydrogen gas =
`(1.65-x)=(1.65-0.05)=1.60mol`

Putting values in above expression, we get:

`K_c=((0.100)^2)/(1.25* (1.60)^3)\\\\K_c=1.95* 10^(-3)`

Calculating the
`K_c'` for the given chemical equation:

`2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)`

`K_c'=(1)/(K_c)\\\\K_c'=(1)/(1.95* 10^(-3))=5.13* 10^2`

Hence, the value of
`K_c` for
`2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)` reaction is
`5.13* 10^2`