Question

A vessel of volume 22.4 dm3 contains 20 mol h2 and 1 mol n2 ad 273.15 k initially. All of the nitrogen reacted with sufficient hydrogen to form nh3. Calculate the total pressure and the partial pressure of each component in the final mixture at 273.15 k.

Answer 1

Nitrogen combine with hydrogen to produce ammonia
`\text{NH}_3` at a
`1:3:2` ratio:

`\text{N}_2 \; (g) + 3 \;  \text{H}_2 \; (g) \leftrightharpoons 2\; \text{NH}_3 \; (g)`

Assuming that the reaction has indeed proceeded to completion- with all nitrogen used up as the question has indicated.
`3 \; \text{mol}` of hydrogen gas would have been consumed while
`2 \; \text{mol}` of ammonia would have been produced. The final mixture would therefore contain

• 
  `17 \; \text{mol}` of
  `\text{H}_2 \; (g)` and
• 
  `2 \; \text{mol}` of
  `\text{NH}_3 \; (g)`

Apply the ideal gas law to find the total pressure inside the container and the respective partial pressure of hydrogen and ammonia:

• 
  `\begin{array}{lll} P(\text{container}) &= & n \cdot R \cdot T / V \\ & = & (17 + 2) \; \text{mol} * 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^(-1) \cdot \text{K}^(-1) \\ & &* 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.926 * 10^(3) \; \text{kPa} \end{array}`
• 
  `\begin{array}{lll} P(\text{H}_2) &= & n \cdot R \cdot T / V \\ & = & (17) \; \text{mol} * 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^(-1) \cdot \text{K}^(-1) \\ & &* 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  1.723 * 10^(3) \; \text{kPa} \end{array}`
• 
  `\begin{array}{lll} P(\text{NH}_3) &= & n \cdot R \cdot T / V \\ & = & (2) \; \text{mol} * 8.314 \; \text{L} \cdot \text{kPa} \cdot \text{mol}^(-1) \cdot \text{K}^(-1) \\ & &* 273.15 \; \text{K} / (22.4 \; \text{L}) \\ &=&  2.037 * 10^(2) \; \text{kPa} \end{array}`