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Problem 14.61

For the water-gas reaction (Problem 43), K_{c} = 0.111 at about 1100 K. If 0.100 mol H_{2}O(*g*) and 0.100 mol H_{2}(*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.

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( |
+ |
H |
CO( |
+ |
H |
||

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.