Answer the following questions about reactions and their equilibrium constants.
ID: 481862 • Letter: A
Question
Answer the following questions about reactions and their equilibrium constants.
Part 1:
Ammonia can be formed from hydrogen and nitrogen gases according to the following reaction:
3H2(g) + N2(g) --> 2NH3(g)
Alternatively, a mole of ammonia can be broken up into hydrogen and nitrogen gases according to the following formula:
NH3(g) --> 3/2 H2(g) + 1/2 N2(g)
What changes have occurred from the first to the second version of the reaction?
A
The reaction has been halved.
B
The reaction has been flipped and halved.
C
The reaction has been flipped.
Part 2:
Which transformations of the equilibrium constant K would you do from the first to the second reaction in part 1?
A
Make K negative.
B
Cut K in half.
C
Take the square root of K.
D
Take the inverse of K.
E
Take the inverse of K (1/K) and take the square root.
F
Cut K in half and make it negative.
Part 3:
At a particular temperature, the equilibrium constant K of the first reaction in part 1 is equal to 5.40. What would the equilibrium constant be for the second reaction in part one?
Part 4:
If the Kc for the first reaction in part one is 5.40, calculate Kp for this reaction. Note, the temperature of this reaction is 297.7 K. (hint: if the pressures of the equilibrium constant Kp should be in atmospheres, what value of R should you use?)
Explanation / Answer
1)
Clearly the reaction has been halfed and reversed
Answer: B
2)
fliping equation inverses the equilibrium constant
multiplying the reaction by a factor a, raises power a to the equilibrium constant
K' = sqrt (1/k)
Answer: E
3)
K' = sqrt (1/K)
= sqrt(1/5.40)
= 0.43
Answer: 0.43
4)
n = 2 - 3 - 1 = -2
use
Kp = Kc*(RT)^n
= 5.40*(0.0821*297.7)^(-2)
= 5.40 * 1.674*10^-3
= 9.04*10^-3
Answer: 9.04*10^-3