Question

The value of equilibrium constant in reaction of phosphorylation is 3.8×10-³ at 298 Kelvin temperature. Calculate the value of ∆G° and give the opinion for the reaction

Answer 1

Hi mate.

Thanks for asking this question,

Here is your answer,



Given :

K = 3.8 × 10-3

T = 298 k

R = 8.314 J/ KMol.



By the relationship between the standard

Gibbs energy and Equilibrium constant is



$d {g}^{o} \: = - rt \: lnk$


$d {g}^{o} = -2.303 \: rt \: \: log_{10}(k)$

$d {g}^{o \:} \: = -2.303 \times 8.314 \times 298 \times \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: log_{10}(0.0038)$


$d {g}^{o} = - 8.314 \times 298 \times - 2.4202 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \times 2.303$


$d {g}^{o} = -2.303 \times 2477.57 \times - 2.4202$

$d {g}^{o} = - 5996.21 \times 2.303$


$d {g}^{o} \: = - 5.99 \times {10}^{3} \times 2.303$

$d {g}^{o} = - 13.79 \times {10}^{3} \: KJ {mol}^{ - 1}$





Thank you.




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