00:01
We're given the rate law for a particular chemical reaction where the rate is equal to the rate constant multiplied by the concentration of n2 raised to the third and the concentration of o3 raised to the 1.
00:17
To determine the order of the reaction with respect to n2, that's just the superscript or the exponent here.
00:25
So it is third order with respect to n2 and first order if there's nothing here that means there's a one with respect to 03 the overall order is just the sum of the two orders for each reactant giving us fourth order overall and then the third question says at a certain temperature we have n2 and 03 at a particular at particular concentrations, and the initial rate is 89 .0 molar per second.
01:10
What would the initial rate be if the concentration of n2 were halved? well, with it being third order with respect to n2, then if we cut the concentration in half, and we raise that to its order, we get one -eighth the original rate.
01:36
So if we take one -eighth, multiplied by the original rate, original rate of 89, then this gives us a new rate of 11 .125, or just 11 .1 1 molar per second.
01:58
And then for the fifth question, the rate of the reaction is measured to be 0 .5, when n2 is 0 .98, and 03 is 1 .1...