Question

# Consider the reaction CO (g) + 0.5 O2 (g) -> CO2 (g). Compute the molar delta...

Consider the reaction CO (g) + 0.5 O2 (g) -> CO2 (g). Compute the molar delta H (in kJ/mol) for this reaction at 298 K and a pressure of 30 bar. Joule-Thomson coefficients and heat capacities are listed in the table below:

 Compound Cp (cal mol^-1 K^-1) Joule-thomson coefficient (K/bar) CO 6.3423 + 0.0018363 T 1.20 O2 6.148 + 0.003102 T 1.15 CO2 6.369 + 0.0101 T 1.10

#### Homework Answers

Answer #1

CO (g) + 0.5 O2 (g) ----------> CO2 (g)

Cp of CO = 6.3423 + 0.0018363 T

Cp of CO } at T = 298 K = 6.3423 + 0.0018363 * 298

= 6.8895 Cal /molK

Similarly,

Cp of O2 } at T = 298 K = 6.148 + 0.003102 * 298

= 7.0724 Cal /molK

Cp of CO2 } at T = 298 K = 6.369 + 0.0101 * 298

= 9.3788 Cal /molK

Using Joule thomson Coefficient for the molecules:

Hf of CO = - 6.8895 * 1.20 = - 8.2674 Cal/mol

Hf of O2 = - 7.0724 * 1.15 = - 8.13326 Cal/mol

Hf of CO2 = - 9.3788 * 1.10 = - 10.31668 Cal/mol

Now,

Hrxn = Hf (CO2) - [Hf (CO) + Hf (O2)]

= - 10.31668 - [- 8.2674 + 0.5*(- 8.13326)]

= 2.017 Cal/mol

= 0.008439 kJ/mol

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