Answered

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Under certain conditions, the equilibrium constant of the reaction below is kc=0. 25. If the reaction begins with a concentration of 0. 0638 m for each of i2 and i− and a concentration of 0. 0 m for i−3, what is the equilibrium concentration of i−?.

Sagot :

It can be deduced that the equilibrium concentration of i− will be 0.053M.

How to calculate the equilibrium concentration

From the information given, the equilibrium constant of the reaction below is kc=0. 25. Since the concentration has also been provided. This will be calculated thus:

[(0.0638 - x)(0.0638 - x)]/x = 0.25

Cross multiply

0.00407 + x² - 0.1276x = 0.25x

x² - 0.3776x + 0.00407 = 0

x = 0.011106M

Therefore, the concentration of I at equilibrium will be:

= 0.0638 - 0.011106

= 0.053M

Therefore, the equilibrium concentration of i− is 0.053M.

Learn more about equilibrium on:

https://brainly.com/question/13414142

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