Roberts & Company Publishers Cmemistry A reaction 4Cag has a standard free-energ

Question

Roberts & Company Publishers Cmemistry A reaction 4Cag has a standard free-energy change of-548 kJ/mol at 25°?. What are the concentrations of A, B, and C at equilibrium i, at the beginning of the reaction, their concentrations are 0, 30 M.0.40 M, and O M. respectively? Number Number Number How would your answers above change if the reaction hud a standatd tree energy change of +5 48 kJmor D.ll There would be more A and B but less Glbi D All concentrations would be higher There would be no change to the answers A? concen rations would be lower. There woue less A and B biz more G

Explanation / Answer

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 = -R?T?ln(K)
<=>
K = e^{ - ?G/(R?T) }

So the equilibrium constant for this reaction at 25°C = 298 K is
K = e^( 5480 J?mol?¹ / ( 8.3144 J?K?¹?mol?¹ ? 298 K) = 9.133

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
9.133 = x /( (0.3 - x)?(0.4 - x) )
<=>
9.133?(0.12 - 0.70?x + x²) = x
<=>
9.133?x² - 7.393?x + 0.936 = 0

The solutions to this quadratic equation are:
x = [ 7.393 ± ?( (7.393)² - 4 ? 9.133 ? 0.936 ) ] / [ 2 ? 9.133 ]
=>
x = 0.157
or
x = 0.652

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

So the equilibrium cocentrations are
[A] = 0.30 - 0.157 = 0.143 M
[B] = 0.40 - 0.157 = 0.243 M
[C] = 0.157 M.

For 2nd part : a reaction with a positive value of DG is reactant-favored and requires the input of energy to go. So Answer ia (A) There would be more A and B and less C.

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