00:01
We take note that the equilibrium constant depends on temperature, but if the reaction is changed, then the equilibrium concept value can also change accordingly.
00:11
For example, a reaction that has, say, reactant forming product is characterized with the equilibrium concept that is equal to k.
00:19
If this reaction is reversed such that we get p forming r, then the equilibrium concept will be the reciprocal of the original equilibrium constant if the reaction original reaction is multiplied by n such that we have nr forming np then the equilibrium constant will be raised to the same n so for this problem we're given with the following reactions we have d forming b given with the equilibrium constant of that is equal to 0 .115 and that we have d forming 2a the equilibrium constant for which is equal to 2 .10.
01:13
We want to find the k value for the reaction described by 2a forming d.
01:26
So to call it more properly, we will use k1 for the first reaction and k2 for the second reaction.
01:34
All right, so let us take note that if we add the two equations given, so let's write that first.
01:40
D forming b and then b forming 2a adding these so we're gonna add d and b we also do the same for the product so you get b plus 2a and simplifying so that we cancel the b we get b forming 2a now the equilibrium constant for this since we add the two reactions we're not gonna add equilibrium constant but rather we'll multiply so it will be k1 times k2 but this is not the reaction that we desire...
