DAT Bootcamp Question

[captainGrape]

14. N2(g) + 3H2(g) ⇌ 2NH3(g) ; ΔH° = -92.22 kJ•mol-1


Which change would decrease the Keq in the above reaction?

Options:

• A. Adding an iron catalyst.
• B. Decreasing the volume of the vessel.
• C. Injecting H2 gas.
• D. Cooling ammonia gas into a liquid.
• E. Increasing the temperature.

I chose that cooling ammonia gas into a liquid would change Keq. I realized after that only temperature can change Keq; but what stumps me is that if we cool NH3 to a liquid the Keq = 1/([N2]([H2]^3)) and that completely changes the equilibrium constant equation; so I don't really understand why this wouldn't be an answer as well.

My reasoning was that 1/[N2][H2]^3 would have a large denominator and would be much smaller than [NH3]^2/[H2]^3[N2]. But not sure and I've tried checking other resources but I can't find anything that specifically addresses this question.

Thanks in advance for the help!

---

Members don't see this ad.

 

[asdf99]

(Goofed the explanation, see OPs response for actual answer)

 

[captainGrape]

Hi! thanks for your answer; just in case you were wondering, or if anyone else was wondering; someone answered my question.

While we cooled ammonia to a liquid, the overall equation for Keq is actually still the same (it still makes ammonia gas); thus, the change into ammonia liquid just makes the equilibrium shift right, which simulates taking products out of solution

 

[asdf99]

Hi! thanks for your answer; just in case you were wondering, or if anyone else was wondering; someone answered my question.

While we cooled ammonia to a liquid, the overall equation for Keq is actually still the same (it still makes ammonia gas); thus, the change into ammonia liquid just makes the equilibrium shift right, which simulates taking products out of solution

Lul thanks for following up . Can't believe I neglected the part about forming more ammonia gas. Fuark , crazy how much you can forget after the dat

Best of luck with the rest of your studying

 

[Bersabeh]

14. N2(g) + 3H2(g) ⇌ 2NH3(g) ; ΔH° = -92.22 kJ•mol-1


Which change would decrease the Keq in the above reaction?

Options:

• A. Adding an iron catalyst.
• B. Decreasing the volume of the vessel.
• C. Injecting H2 gas.
• D. Cooling ammonia gas into a liquid.
• E. Increasing the temperature.

I chose that cooling ammonia gas into a liquid would change Keq. I realized after that only temperature can change Keq; but what stumps me is that if we cool NH3 to a liquid the Keq = 1/([N2]([H2]^3)) and that completely changes the equilibrium constant equation; so I don't really understand why this wouldn't be an answer as well.

My reasoning was that 1/[N2][H2]^3 would have a large denominator and would be much smaller than [NH3]^2/[H2]^3[N2]. But not sure and I've tried checking other resources but I can't find anything that specifically addresses this question.

Thanks in advance for the help!
E is d answer.