​If a 0.1 M solution of glucose 1-phosphate is incubated with a catalytic amount of phosphoglucomtase, the glucose 1-phosphate is transformed to glucose 6-phosphate. At equilibrium, the concentrations of the reaction components are:​

​What would be the calculated values for $$K^{'}_{eq}$$ and $$∆G^0$$ for this reaction at 25°C?​

Correct option is B

To calculate the equilibrium constant (K_eq) and standard free energy change (ΔG°') for the given reaction, we can use the following formulas:

1. Equilibrium constant (K_eq):

We can calculate the equilibrium constant using the formula:

​$$K_{eq} = \frac{[Glucose 6 - phosphate]}{[Glucose 1 - phosphate]}$$​

Given:

• [Glucose6−phosphate]=9.6×10−2M[Glucose 6-phosphate] = 9.6 \times 10^{-2} \, M[Glucose6−phosphate]=9.6×10^-2 M
  
  

• [Glucose1−phosphate]=4.5×10−3M[Glucose 1-phosphate] = 4.5 \times 10^{-3} \, M[Glucose1−phosphate]=4.5×10^-3 M

​$$K_{eq} = \frac{9.6 \times 10^{-2}}{4.5 \times 10^{-3}} = 21.33$$

2. Standard Free Energy Change (ΔG°'):

The standard free energy change (ΔG°') is related to the equilibrium constant (K_eq​) by the following equation:

​$$\Delta G^\circ = -RT \ln K_{eq}$$  

Where:

• R is the gas constant = 8.314 J/(mol·K)

• T is the temperature in Kelvin. Since the problem gives the temperature as 25°C, we need to convert it to Kelvin:
  $T=25+273.15=298.15K$

  ​$$\Delta G^\circ$$ =$$-8.314\times 298.15\times \ln (21.33)$$   

where, 

$$\ln (21.33)\approx 3.063$$

Therefore, the result is:

​$$\Delta G^\circ$$= -7595.2 J/Molar ,$$\Delta G^\circ$$= -7.6 KJ/M

Final Answer:

The calculated values forKeqK_{eq}K^eq and $$\Delta G^\circ$$​G∘\Delta G^\circΔG∘ are:

​K^eq  =21.33 

​$$\Delta G^\circ$$ = -7.6 KJ/M​