Part #1: Ammonia can be formed from hydrogen and nitrogen gases according to the

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)

Part #2:

Which transformations of the equilibrium constant K would you do from the first to the second reaction in part 1?

Part #3:

At a particular temperature, the equilibrium constant K of the first reaction in part 1 is equal to 7.34. 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 7.34, calculate Kp for this reaction. Note, the temperature of this reaction is 292.2 K. (hint: if the pressures of the equilibrium constant Kp should be in atmospheres, what value of R should you use?)

Explanation / Answer

1)

Ka1 = [NH3]^2/[H2]^3[N2]

if you need

Ka2 = [H2]^3[N2]/[NH3]^2

then we must invert it

that is

Ka2 = 1/Ka1

3)

K1 = 7.34

then; for reaction 2

K2 = 1/K1 = 1/7.34 = 0.1362397

4)

You can relate Kp and Kc as follows

Kp = Kc*(RT)^dn

dn = moles of products - moles of reactant

dn = 2-(3+1) = -2

R = 0.082 atmL/molK

T = 292.2 K

then

Kp = Kc*(RT)^dn

Kp = 7.34*(0.082*292.2)^(-2)

Kp = 0.01278521194

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects