For a certain reaction, K⁰ = 1.2 × 104 at 25 °C, and K⁰ = 4.3 × 103 at 125 °C

Question

12. For a certain reaction, K⁰ = 1.2 × 104 at 25 °C, and K⁰ = 4.3 × 103 at 125 °C. (a) Is this reaction endothermic, or is it exothermic? (b) Calculate ΔHr⁰.

14. For the reaction in exercise 12, what is the value of K⁰ at 765 °C?

Answer 1

You can also tell it is exothermic since at higher temps, the K value is lower, suggesting that heat is a product of the reaction.

ln(K2/K1) = -∆H/R (1/T2 - 1T1)

K1 = 1.2x104

K2 = 4.3x103

R = 8.314 J/Kmol

T1 = 298K

T2 = 398K

ln(4.3x103 / 1.2x104) = -∆H/8.314 (1/398 - 1/298)

-1.099 = -∆H/8.314 (0.002513 - 0.003356)

-1.099 = -0.00084∆H/8.314

∆H = 10,877 J/mole = 10.9 kJ/mole

14). Use the above van't Hoff equation to solve for K @ 765C = 1038K

ln(K2/K1) = -∆H/R (1/T2 - 1/T1)

K1 = 1.2x104

K2 = ?

T1 = 298K

T2 = 1038K

∆H = 10,877 J/mol (note to use same units as those for R)

R = 8.314 J/Kmol

Plug and chug and solve for K2