Question

Given the following photosynthesis reaction:

6CO2 (g) + 6H2O (g) -> C6H12O6 (s) + 6O2 (g)

At room temperature. The following equilibrium concentrations were found: [H2O] = 7.91x10^-2M, [CO2] = 9.3^-1​M, and [O2] = 2.4x10^-3M

(1a) What is the equilibrium constant expression, Kc

(1b) Calculate the value of Kc for the reaction at this temperature

(1c) Calculate the value of Kp for the reaction at room temperature:

(1d) If a plant is placed In an atomosphere that has a meassured pressure of 780 torr and contains 10% O2 gass, 40% H2O gass, and 50% CO2 gass, what is the value of Qp?

(1e) In which direction will the reaction procees to reach equilibrium?

(1f) What are the partial pressure of H2O, CO2, and O2 at equilibrium?

[This is all information given to answer questions]

Answer 1

1) 6CO2 (g) + 6H2O (g) -------> C6H12O6 (s) + 6O2 (g)

equilibrium concentration 9.3 10-1​M 7.91x10-2 M 2.4x10-3

Kc = [O2]6 / [CO2]6 [H2O]6

Kc= [  2.4x10-3]⁶ ./ [9.3 10-1]⁶      [7.91x10-2]⁶

= 191.10 * 10^-18 / 646990.18 * 10^{-6 *} 244939 * 10^-12

= 191.10 / 158473127699 =

Kc= 1.206 * 10^-8

3) As Kp= Kc * (RT)ⁿ n = no of moles of product - no of moles of reactant = 6-12 =-6 moles

= 1.206 * 10^-8 ( 0,0821 * 298) ^-6

= 1.206 * 10^-8   ( 24.4658)

= 1.206 * 10^-8 * 4.66 * 10^-9

Kp = 5.6199 * 10^-17

d) Q= [0.1]⁶ / [0.4]⁶   [0.5]⁶

= 1 * 10^-6 / 0.004096 * 0.015625

=   1 * 10^-6 / 0.000064 =  1 * 10^-6 / 64 * 10-6 =

Q = 0.015625

5) As the value of Q > Kc the reaction will shift in the backward direction