has a standard free-energy change of -3 40 kJ/mol at 25 degree C. What are the c

Question

has a standard free-energy change of -3 40 kJ/mol at 25 degree 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? How would your answers above change if the reaction had a standard free-energy change of +3.40 kJ/mol? All concentrations would be higher. There would be less A and B but more C. All concentrations would be lower. There would be no change to the answers There would be more A and B but less C.

Explanation / Answer

The equilibrium concentration 0 for compound A is definitely wrong. The equilibrium concentrations of the components involved in a reaction are never zero. If this would happen, the reaction quotient in equilibrium becomes zero (zero concentration component is a product) or infinite (zero concentration component is a reactant). That means
K = 0
or
K =
if any concentration is zero, and both cases are infeasible.

Its hard to say what went wrong without knowing the reaction equation. I guess the reaction follows the reaction equation
A + B C

Then the concentrations in equilibrium state satisfy the relation:
K = [C]/ ( [A][B] )

Equilibrium constant and free energy change are related as:
G = -RTln(K)
<=>
K = e^{ - G/(RT) }

So the equilibrium constant for this reaction at 25°C = 298 K is
K = e^( 3400 Jmol¹ / ( 8.3144 JK¹mol¹ 298 K) = 3.94

ICE table
.......... [A]......... [B]......... [C]
I.......... 0.3........ 0.4.......... 0
C........ - x.......... - x........ + x
E....... 0.3 - x.... 0.4 - x...... x

When you substitute the expressions for the equilibrium concentrations from the last row to the equilibrium equation you get
3.94 = x /( (0.3 - x)(0.4 - x) )
<=>
3.94(0.12 - 0.50x + x²) = x
<=>
3.94x² - 1.97x + 0.47 = 0

The solutions to this quadratic equation are:
x = [ 1.97 ± ( (1.97)² - 4 3.94 0.47 ) ] / [ 2 3.94 ]
=>
x = 0.011
or
x = 0.488

The second solution is infeasible because it would lead to negative concentrations for A and B.

So the equilibrium cocentrations are
[A] = 0.30 - 0.011 = 0.289 M
[B] = 0.40 - 0.011= 0.389 M
[C] = 0.011 M

There would be less A and B but more C

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