Chemistry Labs

Problem 4

A vessel of volume 5.0 dm3^3 was filled with ethane at 300 K and 101.325 kPa, sealed, and then heated. The measured pressures were: at 300 K: 101.3 kPa (theoretical p′=101.3p'=101.3 kPa); at 500 K: 169.8 kPa (p′=168.7p'=168.7 kPa); at 800 K: 276.1 kPa (p′=269.9p'=269.9 kPa); at 1000 K: 500.7 kPa (p′=337.4p'=337.4 kPa). (a) Explain why the measured pressure exceeds p′p' at higher temperatures and write the reaction equation. (b) Calculate the degree of conversion α\alpha and equilibrium constant KpK_p at 800 K and 1000 K. (c) Using the integrated van 't Hoff equation ln⁡(Kp,2/Kp,1)=−ΔH∘R(1T2−1T1)\ln(K_{p,2}/K_{p,1}) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right), calculate the mean reaction enthalpy ΔH∘\Delta H^\circ between 800 K and 1000 K. (R=8.314R = 8.314 J mol−1^{-1} K−1^{-1})
Step 3 of 4: Equilibrium constants
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Kp=p(CX2HX4) p(HX2)p(CX2HX6)=α21−α2 p  [or α21−α p′]K_p = \dfrac{p(\ce{C2H4})\,p(\ce{H2})}{p(\ce{C2H6})} = \dfrac{\alpha^2}{1 - \alpha^2}\,p \;\left[\text{or }\dfrac{\alpha^2}{1 - \alpha}\,p'\right]
Analysis

Partial pressures: p(CX2HX4)=p(HX2)=α p′p(\ce{C2H4}) = p(\ce{H2}) = \alpha\,p', p(CX2HX6)=(1−α)p′p(\ce{C2H6}) = (1 - \alpha)p'. At 800 K: Kp=0.02320.977×269.93=0.146K_p = \frac{0.023^2}{0.977} \times 269.93 = 0.146 kPa. At 1000 K: Kp=0.48420.516×337.41=153.2K_p = \frac{0.484^2}{0.516} \times 337.41 = 153.2 kPa.