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Question: What is the \({K_{eq}}\) of the following reaction? ![](https://www.vedantu.com/question-sets/296a...

What is the Keq{K_{eq}} of the following reaction?

A. 1.71.7
B. 0.170.17
C. 1717
D. None of these

Explanation

Solution

If the value of change in Gibbs free energy under standard condition is negative then the value of the equilibrium constant will be greater than one which implies that the reaction is spontaneous and the reaction will proceed in forward direction but if the value of change in Gibbs free energy under standard condition is less than zero then the reaction will proceed in reverse direction.

Complete answer:
Step 1: Gathering the values from the question:
According to the question ΔG∘=−0.010 kJ mol−1\Delta {G^\circ } = - 0.010{\text{ kJ mo}}{{\text{l}}^{ - 1}}
The reaction proceeds at room temperature therefore the temperature at Celsius is 27∘C{27^ \circ }C
while the temperature in Kelvin is 273+27= 300K.273 + 27 = {\text{ }}300K.
The value of universal gas constant R in kJ mol−1kJ{\text{ }}mo{l^{ - 1}} is 8.314kJ mol−18.314kJ{\text{ }}mo{l^{ - 1}}.
Step 2: Substituting the values in the equation:
If the value of Gibbs free energy change is known for a chemical process at a known temperature the value of equilibrium constant can be calculated from the equation mentioned below.
ΔG∘=2.303 RT log Keq\Delta {G^\circ } = 2.303{\text{ RT log }}{{\text{K}}_{eq}}
log⁡ Keq=ΔG∘2.303RT log\log {\text{ }}{{\text{K}}_{eq}} = \dfrac{{\Delta {G^\circ }}}{{2.303RT{\text{ log}}}}
log⁡ Keq=−0.0102.303×8.314×300\log {\text{ }}{{\text{K}}_{eq}} = \dfrac{{ - 0.010}}{{2.303 \times 8.314 \times 300}}
Step 3: solving the equation:
log⁡ Keq=−0.01019.147×300\log {\text{ }}{{\text{K}}_{eq}} = \dfrac{{ - 0.010}}{{19.147 \times {\text{300}}}}
log⁡ Keq=−0.0105744.1\log {\text{ }}{{\text{K}}_{eq}} = \dfrac{{ - 0.010}}{{5744.1}}
log⁡ Keq=−1.74×10−6\log {\text{ }}{{\text{K}}_{eq}} = - 1.74 \times {10^{ - 6}}
Keq= 0.999  {K_{eq}} = {\text{ }}0.999\;
Therefore, Keq{K_{eq}} is approximately equal to 1.

Note:
In an exergonic process the energy will be releases while the Gibbs free energy is less than zero while at equilibrium Gibbs free energy is equal to zero and no net changes occur in the system and in endergonic process energy is required and work is done on the system while the Gibbs free energy is more than zero. If the value of Keq{K_{eq}} under standard condition is more than one then the reaction will proceed in forward direction but when the value of Keq{K_{eq}} under standard condition is less than one the reaction will proceed in backward direction.