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lbvjy [14]
3 years ago
15

The δg°\' of the reaction is -"8.550" kj ·mol–1. calculate the equilibrium constant for the reaction. (assume a temperature of 2

5°
c.)
Chemistry
1 answer:
Alex787 [66]3 years ago
5 0

31.5. If Δ<em>G</em>° = -8.550 kJ·mol^(1) at 25 °C, <em>K</em> = 31.5.

The relationship between Δ<em>G</em>° and <em>K</em> is

Δ<em>G</em>° = -<em>RT</em>ln<em>K</em>

where

<em>R</em> =the gas constant = 8.314 J·K^(-1)mol^(-1)

<em>T</em> is the Kelvin temperature

Thus,

ln<em>K</em> = -Δ<em>G</em>°/(<em>RT</em>)

In this problem,

<em>T</em> = (25 + 273.15) K = 298.15 K#

∴ ln<em>K</em> = -[-8550 J·mol^(-1)]/[8.314 J·K^(-1)mol^(-1) x 298.15 K]

= [8550 J·mol^(1)]/[2479 J·mol^(-1)] = 3.449

<em>K</em> = e^3.449 = 31.5

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