Part 1: Ammonia can be formed from hydrogen and nitrogen gases according to the
ID: 483734 • Letter: P
Question
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 .Take the inverse of K (1/K) and take the square root
B. Cut K in Half
C. Take the inverse of K
D. Make K negative
E. Cut K in half and make it negative
F. Square Root K
Part 3:
At a particular temperature, the equilibrium constant K of the first reaction in part 1 is equal to 6.00. 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 6.00, calculate Kp for this reaction. Note, the temperature of this reaction is 284.9 K. (hint: if the pressures of the equilibrium constant Kp should be in atmospheres, what value of R should you use?)
Explanation / Answer
Part 1: 3H2+ N2 --->2NH3
Flip the above reaction we have
2NH3 ----> 3H2 + N2
Equilibrium constant,K= ([H2]^3[N2])/[NH3]^2
On multiply the above reaction with 1/2 we get the required rquation
NH3 -----> (3/2)H2 + (1/2) N2.
Equilibrium constant
,K'= ([H2]^(3/2) Xx[N2]^(1/2) ) /[NH3]
. =1/sq root K
So option (B) is correct
Part 2:
Option A is correct.
Part 3 :
Given K = 6.00
K' =1/K
= 1/6
= 0.44