An equilibrium mixture contains 0.400 mol of each of the products (carbon dioxid

Question

An equilibrium mixture contains 0.400 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.....PLEASE HELP ME WITH THIS QUESTION I DO NOT UNDERSTAND HOW TO WORK IT. Thank you!!!

Question 2 of 27 Map General Chemistry 4th Edition University Science Books e Rock presented by Sapling Leaming An equilibrium mixture contains 0.400 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 CO, H 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? Number mo l

Explanation / Answer

CO (g) + H2O (g) <--> CO2 (g) + H2 (g)

given at equilibrium CO moles = H2O moles = 0.2 , CO2 moles = H2 moles = 0.4

[CO] = 0.2/1 = [H2O] . [CO2] =[H] =0.4/1 = 0.4M

K = [CO2] [H2] / [CO] [H2O] ,

= ( 0.4 x 0.4) / ( 0.2 x 0.2) = 4

Let m be moles of CO2 added , then CO2 moles = 0.4+m

at equilibrium CO2 moles = 0.4+m -0.3 = 0.1+m ( since 0.3 CO2 converted to 0.3 CO)

H2 moles = 0.4-0.3 = 0.1

gven CO moles after CO2 addition = 0.2+0.3 = 0.5 moles = H2O moles

4 = (0.1+m) ( 0.1) / ( 0.5) ( 0.5)

m = 9.9 moles

hence number of CO2 moles needed to be added = 9.9

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