Question 4.6: Calculate the standard heat of the following methanol-synthe...
Calculate the standard heat of the following methanol-synthesis reaction at 800°C:
\mathrm{CO}(g)+2 \mathrm{H}_2(g) \rightarrow \mathrm{CH}_3 \mathrm{OH}(g)
Apply Eq. (4.16) to this reaction for reference temperature T0 = 298.15 K and with heat-of-formation data from Table C.4:
ΔH° =\sum\limits_{i}{ν_{i} ΔH^{°}_{ft}} (4.16)
\Delta H^{°}_{0}=H^{°}_{298}= -200,660 – (−110,525) = − 90,135 J
Evaluation of the parameters in Eq. (4.20) is based on data taken from Table C.1:
\int_{T_{0} }^{T}{\frac{\Delta C^{o}_{P} }{R} dt}=\Delta A (T- T_{0} )+\frac{\Delta B}{2} \left(T^{2} – T^{2}_{0} \right) + \frac{\Delta C}{3} \left(T^{3} – T^{3}_{0} \right)\Delta D\left(\frac{T -T_{0} }{TT_{0} } \right) (4.20)
| i | ν_{ i} | A | 10³B | 10^{6} C | 10^{-5} D | ||
| CH_{3} OH | 1 | 2.211 | 12.216 | − 3.450 | 0.000 | ||
| C O | -1 | 3.376 | 0.557 |
0.000 |
− 0.031 | ||
| H _{2} | -2 | 3.249 | 0.422 | 0.000 | 0.083 |
From its definition,
ΔA = (1)(2.211) + (−1)(3.376) + (− 2)(3.249) = − 7.663
Similarly,
ΔB = 10.815 \times 10^{−3} ΔC = − 3.450 \times 10^{−6} ΔD = − 0.135 \times 10^{5}The value of the integral of Eq. (4.20) for T = 1073.15 K is represented by:
IDCPH ( 298.15, 1073.15; − 7.663, 10.815 \times 10^{−3} , − 3.450 \times 10^{−6} , − 0.135 \times 10^{5} )The value of this integral is −1615.5 K, and by Eq. (4.19),
ΔH° = ΔH^{°}_{0} + R \int_{T_{0} }^{T}{\frac{\Delta C^{°}_{P}}{R}dT } (4.19)
ΔH° = − 90,135 + 8.314 ( −1615.5 ) = −103,566 J