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ale4655 [162]
3 years ago
8

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 h

ydrogen to form nh3. Calculate the total pressure and the partial pressure of each component in the final mixture at 273.15 k.
Chemistry
1 answer:
NikAS [45]3 years ago
6 0

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