00:01
In this problem, we have given that a reacts with 2 moles of p reversibly to give 2 moles of c and d.
00:24
Suppose initial concentration of a is x molar, then we have given that concentration of b is 1 by 5 times of a and c and d are 0.
01:06
At equilibrium, concentration of x is decreased by y, then concentration of b is decreased by 2y, concentration of c is increased by 2y, and concentration of d is increased by y.
01:28
We have given that at equilibrium concentration of a and b are equal.
02:13
So we can write x minus y is equal to 1 .5x minus 2y.
02:23
So we can write 2y minus y is equal to 1 .5x minus x so we get y is equal to 0 .5x minus x so we get y is equal to 0 .5x for this reaction, we have to find the value of kp and we have given information about concentration so we can evaluate kc.
02:57
For this given reaction, delta n which is equal to change in number of moles of gaseous products and gaseous reactants.
03:37
Is equal to number of moles of gases product is 2 plus 1 minus number of moles of gases reactant is equal to 1 plus 2 so this is 0 so here delta n is equal to 0 so kp is equal to kc now we can calculate k c to find the value of kp for this reaction now at equilibrium, concentration of a is equal to x minus y equals to x minus 0 .5x, which is equal to 0 .5x...