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Week 9 Chem 14A Learning Session Number 8 & 1

Posted: Sat Dec 09, 2017 1:10 pm
by Leah Savage 2F
In number 8, it says, will the equilibrium constant K increase, decrease, or remain constant after the following changes to the exothermic reaction 4A(g) + 3B(g) ⇌ 2C(g)?
When the stoichiometric coefficients are doubled, it says increase or decrease, but I thought that it would definitely decrease? Why could it increase or decrease? Wouldn't it decrease no matter the equilibrium concentrations? Also, if the temperature is doubled, does the K decrease because the equilibrium shifts left because it's exothermic? Why does this change the actual K?

Also, in number 1, since it said X was the equilibrium concentration for SO2, I thought that meant that 2x was equal to x, but then it said O2 also had a concentration of 2x, which doesn't really make sense to me. Can someone explain how to solve?


Re: Week 9 Chem 14A Learning Session Number 8 & 1

Posted: Sat Dec 09, 2017 4:44 pm
by Chem_Mod
Hi Leah,

For number 8 when you double stoichiometric coefficients, you double all the exponents in the equilibrium expression. SO, working out the math, you would get that the new K is the old K squared. However, squaring a number can either increase or decrease it. 1/4 squared will result in 1/16 (overall decrease) whereas 4 squared will result in 16 (overall increase). What you do know for certain is that the new K is the old K squared.

Increasing temperature of an exothermic reaction will favor the reverse reaction. Heat is required for the reverse reaction, and released in the forward reaction. Recall that the only way to change K for a given reaction is by changing temperature.

Number 1 has an error that I did not catch until it was recently brought to my attention. The question should read "...x is the equilibrium concentration of O2(g)" rather than SO2. Sorry about that one and good job paying attention.