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Topic: Equilibrium question  (Read 190 times)

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Offline recurringbox

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Equilibrium question
« on: October 08, 2020, 08:02:25 PM »
This is my first time posting so I'm sorry if this is kind of confusing. Here's the question that I'm kind of struggling with:

2SO2 + O2 <=> 2SO3 (these are all in their gaseous states)
Equilibrium constant (K) = 280 at 1000 K

a) Predict the sign of the standard entropy change of the forward reaction.

I said that the sign should be negative because the forward reaction is going from three moles of gas to two moles, which decreases the disorder.

b) Calculate the standard Gibbs free energy change in kJ at 1000 K.

I used ΔG = -RTlnK = -(8.31)(1000)ln(280) and got -46.8 kJ.

c) Predict, giving your reasons, whether the forward reaction is endothermic or exothermic. Use your answers from above.

I'm not sure what the answer is, I think it's exothermic because we learned in class that if the disorder is low/negative, and the reaction is spontaneous (Gibbs free energy is negative), then the reaction should be exothermic at low temperatures? If this is correct, then how should I explain this, or would what I wrote above be good enough?

d) 0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a 1.00 dm flask at 1000 K. Predict the direction of the reaction showing your working.

I'm not sure where to start with this. Should I make an ICE table, find what the final equilibrium concentrations are and find the equilibrium contstant using K = [sulfur trioxide]2/[sulfur dioxide]2[oxygen]?

Or does the equilibrium constant (280) that was given earlier still apply here, and I should just find the reaction quotient using the same equation above and compare it to the already known K value to predict the direction of the reaction instead?


Offline Meter

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Re: Equilibrium question
« Reply #1 on: October 11, 2020, 02:31:21 PM »

I agree with your reasoning.


Remember that the gas constant is usually given in J/mol*K. Remember to convert to kJ/molK or your result will be off by several magnitudes.


Recall ΔG = ΔH - TΔS

If ΔS is negative, T is fairly large, and ΔG is still negative, what can you reasonably predict about the sign of ΔH? 


The equilibrium constant is always constant regardless of initial conditions, as long as the temperature is constant. That's what makes it powerful! Does that help?

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