Delta G and Equilibrium Related Q

Question

If the hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase has a ΔG′∘ of −11.386 kJ/mol at 25 °C, what percentage of substrate remains once the reaction reaches equilibrium assuming no product was initially present? (Round answer to the nearest whole number) (R = 8.315 J/mol · K)

Answer 1

The equilibrium constant for the reaction can be calculated using the following equation:

```

K = e^(-ΔG′∘/RT)

```

where:

* K is the equilibrium constant

* ΔG′∘ is the standard Gibbs free energy change

* R is the gas constant (8.315 J/mol · K)

* T is the temperature (25 °C = 298 K)

Substituting the known values into the equation, we get:

```

K = e^(-(-11.386 kJ/mol)/(8.315 J/mol · K)(298 K))

```

```

K = 1.17 × 10^5

```

The equilibrium constant tells us the ratio of the concentrations of the products to the reactants at equilibrium. Since no product was initially present, the concentration of the product will be equal to the equilibrium constant. The concentration of the substrate will be equal to 1 M minus the concentration of the product.

```

[product] = K = 1.17 × 10^5

[substrate] = 1 M - 1.17 × 10^5 = 1 - 1.17 × 10^5 = 9.99 × 10^-5

```

The percentage of substrate that remains can be calculated as follows:

```

% substrate = ([substrate] / [initial substrate]) * 100%

```

```

% substrate = (9.99 × 10^-5 / 1 M) * 100%

```

```

% substrate = 0.0999% = 0.1% (rounded to the nearest whole number)

```

Therefore, 0.1% of the substrate remains once the reaction reaches equilibrium.