If the reaction SO2(g) + ½O2(g) SO3 (g) has the equilibrium constant Kc = 56, then what is the value of Kc for the following reaction?
2SO3(g) 2SO2(g) + O2(g)
A. 3.2 × 10-4
B. -112
C. 8.9 × 10-3
D. 56
The answer is A
Bob dude I swear
SO2(g) + ½O2(g) ==> SO3 (g) Kc = 56
SO3(g) ==> SO2(g) + ½O2(g) Kc = 1/56
2SO2(g ==> 2SO2(g) + O2(g) Kc = (1/56)^2
If you will write out the Kc expressions for 1, 2, and 3 above you will see where the reciprocal and reciprocal squared come in.
To find the value of Kc for the given reaction, we can use the relationship between the equilibrium constants of forward and reverse reactions.
Given reaction:
2SO3(g) 2SO2(g) + O2(g)
The given reaction is the reverse of the first reaction:
SO2(g) + ½O2(g) SO3(g)
The equilibrium constant (Kc) for the reverse reaction is the reciprocal of the equilibrium constant (Kc) for the first reaction.
Therefore, Kc for the given reaction is the reciprocal of 56:
Kc = 1/56
To calculate the numerical value of Kc, divide 1 by 56:
Kc = 0.0179 (approximately)
Comparing this value to the available options, we can see that the closest option is:
C. 8.9 × 10-3
Therefore, the value of Kc for the given reaction is 8.9 × 10-3.