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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry, 4th edition.

Problem 14.61

For the water-gas reaction (Problem 43), Kc = 0.111 at about 1100 K. If 0.100 mol H2O(g) and 0.100 mol H2(g) are mixed with excess C(s) at this temperature and equilibrium is established, how many moles of CO(g) will be present? No CO(g) is present initially.






Answer:

A typical equilibrium problem: write the reaction, write the mass action expression, set up a table of concentrations, then plug into the mass action expression and solve. Assume a 1.00 L reaction vessel.

 

C(s)

+

H2O(g)

CO(g)

+

H2(g)

 

Initial

xs

 

0.100

 

0

 

0.100

Change

   

– x

 

+ x

 

+ x

Equilibrium

   

0.100 – x

 

x

 

0.100 + x

This is solved using the quadratic equation to give

x = 0.0436 or x = –0.255

Since x must be a positive number x = [CO] = 0.0436 M = 0.0436 mol in a 1.00 L vessel.

Note: because n = +1 for this reaction, the number of moles of CO present at equilibrium depends upon the size of the reaction vessel. For example, if you do the problem assuming a 2.00 L vessel, the amount of CO present at equilibrium is 0.0584 mol.