Consider the reaction below. 3.2 mol of A and 6.7 mol of B are added to a 7 L container. At equilibrium, the concentration of A is 0.187 M. What is the value of the equilibrium constant?
1 A + 2 B ↔ 5 C
The given reaction is
The initial concentrations of A and B can be calculated by dividing the number of moles of A and B by the volume of the container assuming they are gases.
Hence,
Now, we can create an ICE table to calculate the equilibrium concentrations of the molecules,
Initial, M | 0.457 | 0.957 | 0 |
Change, M | -x | -2x | +5x |
Equilibrium, M | 0.457-x | 0.957-2x | 5x |
Note that according to the balanced equation, if x moles of A and 2x moles of B are consumed we get 5x moles of C.
Hence, from the ICE table, we can write the expression of equilibrium constant Kc as follows:
We are given that the equilibrium concentration of A is 0.187 M.
Hence, from the ICE table, we can write
Hence, the value of Kc can be calculated as
Hence, the equilibrium constant of the reaction is 138 approximately. (rounded to three significant figures).
Consider the reaction below. 3.2 mol of A and 6.7 mol of B are added to...
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