Question

At a certain temperature, 0.3411 0.3411 mol of N 2 N2 and 1.661 1.661 mol of H 2 H2 are placed in a 2.50 2.50 L container. N 2 ( g ) 3 H 2 ( g ) − ⇀ ↽ − 2 NH 3 ( g ) N2(g) 3H2(g)↽−−⇀2NH3(g) At equilibrium, 0.2001 0.2001 mol of N 2 N2 is present. Calculate the equilibrium constant, K c Kc .

Answer 1

Answer: The equilibrium constant for the above reaction is 1.31

Step-by-step explanation:

We are given:

Initial moles of nitrogen gas = 0.3411 moles

Initial moles of hydrogen gas = 1.661 moles

Equilibrium moles of nitrogen gas = 0.2001 moles

For the given chemical reaction:

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

Initial: 0.3411 1.661

At eqllm: 0.3411-x 1.661-3x 2x

Evaluating for 'x', we get:

`\Rightarrow (0.3411-x)=0.2001\\\\\Rightarrow x=0.3411-0.2001=0.141`

Volume of the container = 2.50 L

The expression of
`K_c` for the above equation follows:

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

We are given:

`[NH_3]=(2* 0.141)/(2.50)=0.1128M`

`[N_2]=(0.2001)/(2.5)=0.08004M`

`[H_2]=(1.661-(3* 0.141))/(2.5)=0.4952M`

Putting values in above expression, we get:

`K_c=((0.1128)^2)/(0.08004* (0.4952)^3)\\\\K_c=1.31`

Hence, the equilibrium constant for the above reaction is 1.31