Answer:
K =78
Explanation:
Step 1: Data given
2SO3(g) <--> 2SO2(g) + O2(g) kc = 2.3 x 10^-7
2NO3(g) <--> 2NO2(g) + )2(g) kc = 1.4 x 10^-3
Step 2: Calculate K
Lets write out the two reactions in the proper order and look at how they sum together:
2 SO2(g) + O2(g) <---> 2 SO3(g) (1)
2NO3(g) <---> 2 NO2(g) + O2(g) (2)
The two reactions as now written give us the correct reactants and products on the correct sides of the reaction arrow.
Since we have O2 as both a reactant and a product, we can cancel O2 and are not part of the final overall reaction equation.
Koverall = Kc1 * Kc2
Because we reversed reaction number 1 this affects its Kc via the following:
Krev = 1/Kfwd.
We then replace Kc1 with its value for the reverse direction.
So Koverall now = (1/Kfwd) * Kc2
The sum of the two reactions above gives us:
2 SO2(g) + 2 NO3(g) <---> 2 SO3(g) + 2 NO(g)
The problem states to give the K value for the reaction where all the numbers in front of the molecules are (1), and we have (2)'s. So basically if we multiply the whole reaction by 1/2 we'll get the final overall equation we want.
1/2 ( 2 SO2(g) + 2 NO3(g) <----> 2 SO3(g) + 2 NO(g) )
So Kfinal = (Koverall)^1/2
K = ( 1/Kfwd * Kc2)^1/2
K = ( [1 / 2.3 * 10^-7] * 1.4 * 10^-3)^1/2
K =78
