Answer – We are given,
Reaction with standard free energy –
The initial concentration
[A] = 0.30 M , [B] = 0.40 M, [C] = 0.0 M , T = 25+273.15 = 298.15 K
First, we need to calculate the equilibrium constant from the given standard free energy and we know the relationship between these two is as follows:
We need the standard free energy in J, hence the conversion kJ to J is as follows
1 kJ = 1000 J
-3.06 kJ = -3060 J
By plugging the values in the above formula
Now antiln from both side
Now we need to put the ICE chat
A |
B |
C |
|
I |
0.30 |
0.40 |
0.0 |
C |
-x |
-x |
+x |
E |
0.30-x |
0.40-x |
+x |
Now equilibrium expression
By solving the quadratic equation, the value of x is 0.141
Hence the concentration of A, B, and C are as follows :
[A] = 0.30-x
= 0.30 – 0.141
= 0.159 M
[B] = 0.40-x
= 0.40 – 0.141
= 0.259 M
[C] = x = 0.141 M
When we take the standard free energy is positive then the value of the equilibrium constant is less than 1 and the reaction is favored toward the reactant side means the left side.
As the reaction favored toward the reactant side then the concentration of reactant A and B are more than product A at equilibrium.
The correct option is – There would be more A and B but less C.
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