An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

Question

Chemistry

Wags

7 November, 08:58

An equilibrium mixture contains 0.600 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the reactants (carbon monoxide and water vapor) in a 1.00-L container. How many moles of carbon dioxide would have to be added at constant temperature and volume to increase the amount of carbon monoxide to 0.300 mol once equilibrium has been reestablished?

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Salty · Answer 1

First you'll have to solve for the reaction equilibrium contrast, k. I believe that k=[Products]/[reactants], or 0.6^2/0.2^2 = 9

Now solve for the new concentration of CO2, X. 9=0.6*X / (0.3*0.2). This gives a CO2 concentration of 0.9 mol/L, meaning that 0.3 moles of CO2 have been added