00:01
In this problem, we have given that if the equilibrium constant for the reaction, a reacts with 2b reversively to give c and 2 .5 moles of d has equilibrium constant k is equal to 4 .0.
00:31
Then we have to find the value of equilibrium constant for the reaction 2c plus 5 moles of d reacts reversively to give 2 moles of a and 4 moles of d.
00:48
Suppose equilibrium constant for this reaction is denoted by k -daz.
01:18
Now we can write k equals to concentration of c into concentration of d raised to the power 5 by 2, divide by concentration of a into concentration of b raised to the power 2.
01:48
We have given k is equal to 4, which is equal to concentration of c, concentration of d raised to the power 5 by 2 divide by concentration of a into concentration of b raised to the power squaring both side we get 4 raised to the power 2 equals to concentration of c raise to the power 2, concentration of d to the power 5, divide by concentration of a to the power 2 into concentration of b raised to the power 4.
02:48
So we get 16 is equal to c raise to the power 2, d raise to the power 5, a raise to the power 2 into b raised to the power 4...