A reaction: A(aq)+B(aq)<-->C(aq)
has a standard free-energy change of
–4.69 kJ/mol at 25 °C.
What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?
I know you have to calculate K. So I did:
dG = -RT lnK
-4.69=-0.008314x298.15xln(K)
-1.893=ln(K)
e^-1893=K
K=.150
However, I don't know what to do afterwards.
Thanks in advance!
3 answers
I thought you would just set up an ICE table and subtract .15 from the reactants and add .15 to the product but that is not right.
I figured it out. I got the wrong K value too.
You have to take e^-(-1.893)
so e^(1.893)=6.63
From there,
K=(x)/(.3-x)(.4-x) where K=6.63
Solve for x and you get 0.178
To find the values, plug in 0.178 for x to find values at equilibrium.
You have to take e^-(-1.893)
so e^(1.893)=6.63
From there,
K=(x)/(.3-x)(.4-x) where K=6.63
Solve for x and you get 0.178
To find the values, plug in 0.178 for x to find values at equilibrium.
You got it. Great! Good work.
1 answer