Question

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a flask with of ammonia gas and of oxygen gas, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be . Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture.

Answer 1

The question is incomplete, here is the complete question:

Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor, and in the liquid form it is easily transported. An industrial chemist studying this reaction fills a 5.0 L flask with 2.2 atm of ammonia gas and 2.4 atm of oxygen gas at 44.0°C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of nitrogen gas to be 0.99 atm.

Calculate the pressure equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

Answer: The pressure equilibrium constant for the reaction is 32908.46

Explanation:

We are given

Initial partial pressure of ammonia = 2.2 atm

Initial partial pressure of oxygen gas = 2.4 atm

Equilibrium partial pressure of nitrogen gas = 0.99 atm

The chemical equation for the reaction of ammonia and oxygen gas follows:

                    $4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

Initial:               2.2          2.4

At eqllm:        2.2-4x      2.4-3x         2x        6x

Evaluating the value of 'x':  

$\Rightarrow 2x=0.99\\\\x=0.495$

So, equilibrium partial pressure of ammonia = (2.2 - 4x) = [2.2 - 4(0.495)] = 0.22 atm

Equilibrium partial pressure of oxygen gas = (2.4 - 3x) = [2.4 - 3(0.495)] = 0.915 atm

Equilibrium partial pressure of water vapor = 6x = (6 × 0.495) = 1.98 atm

The expression of $K_p$ for above equation follows:

$K_p=\frac{(p_{N_2})^2\times (p_{H_2O})^6}{(p_{NH_3})^4\times (p_{O_2})^3}$  

Putting values in above equation, we get:

$K_p=\frac{(0.99)^2\times (1.98)^6}{(0.22)^4\times (0.915)^3}\\\\K_p=32908.46$

Hence, the pressure equilibrium constant for the reaction is 32908.46