The reaction is glucose with adenosine triphosphate (ATP) to form glucose 6- phosphate and adenosine diphosphate...

The reaction is glucose with adenosine triphosphate (ATP) to form glucose 6- phosphate and adenosine diphosphate (ADP).

This reaction has a calculated equilibrium constant (K) of 1.90 × 105. What are the equilibrium concentrations of glucose and ATP if the initial concentrations of both are 0.100 M?

Expert Solution

Glucose + ATP -----------> glucose 6- phosphate + ADP K = 1.9 x10⁵

Initial 0.1 M 0.1 M 0 0

At equilibrium 0.1 -x 0.1 -x x x

K = x.x / (0.1-x) (0.1-x)

1.9 x10⁵ = x² / (0.1-x)²

435.88 = x /0.1-x

43.588 - 435.88 x = x

436.88 x = 43.588

x = 0.09977 M

Therefore, equilibrium concentrations are

glucose = 0.1 -x = 0.1 - 0.09977 = 0.00023 M

ATP = 0.1 -x = 0.1 - 0.09977 = 0.00023 M

queen_honey_blossom answered 2 years ago

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