The equilibrium constant based on concentration is Kc = 190 for the reaction of carbon monoxide with hydrogen gas at 1000K. What can you infer about the relative rates of the forward and reverse reactions when the gases have the following concentrations at this temperature?
[CO] = 1.7 M
[H2] = 0.83 M
[CH4] = 1.4 M
[H2O] = 0.82 M
~~~
Based on the given concentrations, I calculated that Q is about 1.2, but where do I go from there? If I did it right, Q is much, much smaller than the value of Kc they gave me, so does that mean the forward reaction rate will be much higher than the reverse?
Thanks!
CO + 3H2 ==> CH4 + H2O
Do I have the reaction products OK?
Kc = 190 - (CH4)((H2O)/(CO)(H2)^3
Qc = ? = (1.4)(0.82)^3/1.7)(0.83) = and that gives me about 0.5 and not 1.2. Check my work to make sure that is right. So the products are too small, the reactants are too large and your conclusion is correct that the rate of the forward reaction rate will be larger than the reverse reaction rate.
Just a quick question - why is the 0.82 cubed rather than the 0.83? Your equation matches mine, so wouldn't the concentration of hydrogen, the 0.83, be cubed? And then the concentration of water would be to the first power . . . ?
I did this:
(1.4 x 0.82) / (0.83^3 x 1.7)
With significant figures taken into account, that gave me the 1.2.
You're right. I got mixed up on my numbers. Yes, it's the (H2) that is cubed. I didn't go through the math but your 1.2 probably is correct. We came to the same conclusion because both of our Qc values were less than 190. Thanks for catching that. My eyes just aren't what they used to me.
To determine the relative rates of the forward and reverse reactions, you can compare the value of the reaction quotient, Q, to the equilibrium constant, Kc.
If Q is much smaller than Kc, it means that the concentrations of the reactants are much lower than the concentrations at equilibrium. In this case, the forward reaction is favored and the forward reaction rate is expected to be higher than the reverse reaction rate.
Since Q is calculated to be about 1.2, and Kc is given as 190, we can conclude that Q is indeed much smaller than Kc. Therefore, the forward reaction rate will be much higher than the reverse reaction rate.
Based on the provided concentrations, it is important to note that the reaction quotient, Q, is calculated from the molar concentrations of the reactants and products at a specific point in time, while the equilibrium constant, Kc, represents the ratio of the concentrations at equilibrium.