We are going to answer a question and ask a question in each posting. First you answer the previous post's question and then you ask your own question. Let's try to keep it mostly conceptual (some math allowed) so that we can get at what these concepts mean.
A dynamic equilibrium arises when the rate of a forward reaction becomes equal to the rate of the reverse reaction. This is a process that occurs very often in chemical reactions. The ratio of the rate constant for the forward reaction, kf to the rate constant for the reverse reaction, kr is called an equilibrium constant, K. Thus K = kf /kr.
If the rate constant for the forward reaction equals 1.2x10³ and the rate constant for the reverse reaction equals 4.1x10², what would be the value for the equilibrium constant? What do you think this means in terms of whether there are more reactants or more products at equilibrium?
K = 1.2e3/4.1e2 = 2.927, which falls into the intermediate range, which would indicate that there are roughly an equal amount of reactants and products being produced once dynamic equilibrium is reached.
ReplyDeleteUsing the relationships between Q and K with an equilibrium constant of kc=1.78,
is the following reaction at equilibrium if all of the concentrations are 2? If not what direction will the reaction proceed?
2H2O <--> O2 + 2H2
I'm not entirely sure if you gave enough information in that question or you worded it in a weird way. Because if you meant to use Kp = Kc(RT)^n you didn't give a temperature. The Kc would equal [2]x[2]^2 / [2]^2 meaning Kc = 2. And that's in a intermediate range because it's so close to the equilibrium constant so there are significant amounts of both reactants and products.
ReplyDeleteWhat is the equilibrium constant expression for the following reaction below?
4NO2(g) + O2(g) <--> 2N2O5(g)
For Dustin's question, you would use the balanced equation to write a Q expression and plug in the concentrations that were given as 2 M each. Thus Q = 2 x 2^2/ 2^2 = 2 (not K). Then compare Q of 2 to the 1.78 which means that Q is too big and the reaction will make reactants to get back to equilibrium.
DeleteOh okay, so I was doing the right thing I just got confused on which equations to use, I was basically making it more complicated. That makes sense now, thank you!
DeleteAnswering Emily question:
ReplyDeleteK=((N2O5)^2)/((NO2^4)*(O2))
What is the Kp of the following reaction
2 SO2(g) + O2(g) <--> 2 SO3(g)
At 500K and a Kc of 5.0*10^3
Thanks for your responses!
ReplyDelete