PLEASE SHOW THE SOLUTIONS! THANK YOU SO MUCH! A + B -> C (reversible reaction) 1

Question

PLEASE SHOW THE SOLUTIONS! THANK YOU SO MUCH!

A + B -> C (reversible reaction)

1. For the reversible reaction above, delta G naught = -5kcal/mol at 37 celcius degree. For each of the following statements, indicate whether it is true, false, or there is insufficient information to determine.

(a)The free energy of a solution containing 1M A and 1M B is greater than the free energy of a 1M C solution by 5kcal/mol.

(b) For the forward reaction (as written), A and B are the reactants and C is the product.

(c) The forward reaction will be spontaneous at physiological temperature when A,B, and C are at their physiological concentrations.

(d) You cannot know the equilibrium concentrations of the components without additional information.

(e) You cannot calculate the equilibrium constant at 37 celcius degree from the information provided.

(f) If delta G naught = 0(instead of -5kcal/mol) for the above reaction, then combining 1M A, 1M b, and 1M C would result in no net conversion of reactants to products (or vice versa).

(g) For the example in part (f), the equilibrium constant would be 1.0.

2. For the reaction of problem 1:

(a) Calculate the equilibrium constant at 37celcius degree.

(b) Calculate delta G for the reaction when: [A]=0.1M, [B]=0.1M,[C]=0.05M.

(c) Calculate the equilibrium concentrations of A,B, and C given the initial conditions of part (b).

Explanation / Answer

1)

a)true

b)true

c)true

d)false

e)false

f)true

g)true

2)

a)delta G = - 2.303RT logK

   - 5000 cal = - 2.303x 2x310 logK

K = 3213.66

b) Keq = [c]/[A][B]

      Keq = 0.05/(0.1)(0.1)

       Keq = 5

delta G = - 2.303RT logK

               = - 2.303*2*310 log5

                = - 998.03 cal

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