Question

calculate the equilibrium constant for the reaction at ph 7.0 and 25°c (δg°′ = −20.9 −1) glucose-1-phosphate h2o → glucose h2po4

Answer 1

The equilibrium constant for the reaction at pH 7.0 and 25°C with a given ΔG°′ of -20.9 kJ/mol is approximately 2.09 × 10⁻⁴.

Step-by-step explanation:

To calculate the equilibrium constant (K) for the reaction given the standard free-energy change (ΔG°′), we use the relationship between ΔG° and K which is given by the following equation:

ΔG°′ = -RT ln(K)

Where ΔG°′ is the standard free-energy change, R is the universal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant. To find K, we rearrange the equation:

K = e^{-ΔG°′ / (RT)}

First, we convert the temperature from degrees Celsius to Kelvin:

T = 25°C + 273.15 = 298.15 K

Next, we plug in the values for ΔG°′, R, and T into the equation to calculate K:

K = e{⁽⁻²⁰.⁹ × ¹⁰³ ᴶ/ᵐᵒˡ⁾/ ⁽⁸.³¹⁴ ᴶ/⁽ᵐᵒˡ·ᴷ⁾ × ²⁹⁸.¹⁵ ᴷ⁾}

K = e{⁽⁻²⁰.⁹ × ¹⁰³⁾ / ⁽⁸.³¹⁴ × ²⁹⁸.¹⁵⁾}

K = e{⁽⁻²⁰.⁹ × ¹⁰³⁾ / ⁽²⁴⁷³.⁹⁹⁾}

K = e{⁻⁸.⁴⁵}

K = 0.000209 ~ 2.09 × 10⁻⁴

The equilibrium constant for this reaction at pH 7.0 and 25°C is approximately 2.09 × 10⁻⁴