Question

A closed 1L chamber of gas undergoes the following reaction at 400 °C and 20,000 kPa: 2H2(g)+ O2(g) --> 2H2O(g) Assuming 2 mols of hydrogen gas, 1 mol of oxygen gas, a constant temperature, and no product at start, what is the resulting pressure after the reaction occurs?

Answer 1

The resulting pressure after the reaction occurs is 13,333.3333 kPa

Step-by-step explanation:

The combined gas law for an ideal gas is:

`\frac {P_1* V_1}{n_1* T_1}=\frac {P_2* V_2}{n_2* T_2}`

Where,

P₁ , V₁ , n₁ , T₁ are the pressure, volume, moles and temperature respectively of ideal gas 1.

P₂ , V₂ , n₂ , T₂ are the pressure, volume, moles and temperature respectively of ideal gas 2.

For the question, temperature stays constant and the volume ("closed chamber") are constant.

Thus, equation using for the question:

`\frac {P_1}{n_1}=\frac {P_2}{n_2}`

On the reactant side, number of moles = 2 + 1 = 3 moles

On the product side, number of moles = 2 moles

Given: P₁ = 20,000 kPa

Substituting the values and solving for P₂ gives

20,000kPa / 3 = P₂/2

P₂ = 13,333.3333 kPa