00:01
So this problem gives us several equations as well as their kc equilibrium constants and wants us to find their kp.
00:10
And so first we can write the relation between these.
00:13
So kp equals kc times rt, where r is the gas constant and t is the temperature, to the delta n, which is the number of moles of product minus the number of moles of reactant.
00:28
And so we'll start with letter a.
00:31
And the chemical equation is n204 gas, reversibly forms 2 no2 gas.
00:41
And it gives us a kc of 5 .9 times 10 to the minus third and a temperature of 298 kelvin.
00:51
And so we can plug numbers into our kp equation.
00:54
So we have 5 .9 times 10 to the minus third.
00:58
And then our r value is 0 .0821 times our temperature 298.
01:06
And then in this case we have two moles of product and one mole of reactant.
01:11
So 2 minus 1 is 1, so it's just this to the first.
01:14
And when we solve this, we get that our kp is 0 .144.
01:21
So now moving on to letter b.
01:25
This time we have an equation n2 gas plus.
01:33
3h2 gas reversibly forms 2 an h3 gas.
01:40
We have a kc of 3 .7 times 10 to the eighth and again a temperature of 298 kelvin...