Question

Problem PageQuestion Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 75.0L tank with 23. mol of ammonia gas at 48.°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 21. mol. Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer 1

Answer: The concentration equilibrium constant for the given reaction is 0.14

Explanation:

The molarity of solution is calculated by using the equation:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}$

• For ammonia:

Initial moles of ammonia gas = 23. moles

Volume of the container = 75.0 L

$\text{Initial molarity of ammonia gas}=\frac{23}{75}=0.3067M$

• For hydrogen gas:

Equilibrium moles of hydrogen gas = 21. moles

Volume of the container = 75.0 L

$\text{Equilibrium molarity of hydrogen gas}=\frac{21}{75}=0.28M$

The given chemical equation follows:

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

Initial:                  0.3067

At eqllm:           0.3067-2x       3x         x

Evaluating the value of 'x'

$\Rightarrow 3x=0.28\\\\x=\frac{0.28}{3}=0.0933$

So, equilibrium concentration of ammonia gas = $(0.3067-2x)=[0.3067-(2\times 0.0933)=0.1201M$

Equilibrium concentration of nitrogen gas = x = 0.0933 M

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2]^3[N_2]}{[NH_3]^2}$

Putting values in above equation, we get:

$K_c=\frac{(0.28)^3\times 0.0933}{(0.1201)^2}\\\\K_c=0.14$

Hence, the concentration equilibrium constant for the given reaction is 0.14