Question

1) the photosynthetic conversion of co2 to o2 can be represented by 6co2 + 6h2o <>c6h12o6+6o2
what is the equilibrium expression for this reaction?
2) what is the state of the system if I2=1.0 M and I=1.0×10^-3 M
I2 <>2I. kc=3.8×10^-5
a. the system is at equilibrium
b. the system is at a steady state
c. the system is not at equilibrium and will produce more reactants
d. the system is not at equilibrium and will produce more products
3) what is the effect of increasing the temp on the ammonia formation reaction. n2+3h2 <>2nh3 delta h=-92.38k
a. the equilibrium shifts to the right and k will decrease
b. the equilibrium shifts to the right and k will increase
c. the equilibrium shifts to tell left and k will decrease
d. the equilibrium shifts to the left and k will increase

Answer 1

1.

6 CO₂ + 6 H₂O $\rightleftharpoons$ C₆H₁₂O₆ + 6O₂

K = { [C₆H₁₂O₆] [O₂]⁶ } / { [CO₂]⁶ [H₂O]⁶ }

2.

I₂  $\rightleftharpoons$  2I

I₂ = 1.0 M, I = 1.0 x 10^-3 M

K = [I]² / [I₂]

= (1.0 x 10^-3)² / 1

= 1.0 x 10^-6

But

Kc = 3.8 x 10^-5

= 38 x 10^-6

Hence, [I] will be more and [I₂] will be less.

Answer is (d). The system is not at equilibrium and will produce more products.

3.

N₂ + 3H₂$\rightleftharpoons$  2NH₃$\Delta$H = -92.38 K

Its an exothermic reaction in forward reaction and hence endothermic reaction in reverse reaction.

K = [NH₃]² / { [N₂] [H₂]³ }

Raising the temperature favors the endothermic reaction.

Lowering the temperature favors the exothermic reaction.

Answer is (c). The equilibrium shifts to tell left and k will decrease