1 in each case write the rate expression for the stated reactuon and find the rates of all other

a. The balanced reaction is 6CO2(g) + 6H2O(g) -> C6H12O6(g) + 6O2(g). The rate expression for th

1 in each case write the rate expression for the stated...

In each case, write the rate expression for the stated reaction and find the rates of all other species using the given rate (balance the reactions). a. CO2(g) + H2O(g) -> C6H12O6(g) + O2(g) Rate of formation of O2 = 0.802 M/s b. O3(g) -> O2(g) Rate of depletion of O3 = 2.3x10^-3 M/s c. H2O2(g) -> H2O(g) + O2(g) Rate of formation of H2O = 1.21 M/s

Transcript

00:01 All right, we're giving these three reactions listed here and we're told to balance the reactions as well as find the expression for the rate and the rate of loss or formation of each of these species given some information about one of their species within the reaction.

00:19 So another rate of 02 formation in this reaction is 0 .802.

00:25 So from that, we should be able to derive the rate of loss or production of every species in that reaction.

00:31 So the first thing i'm going to do is go through and balance each of these three equations.

00:35 Because without a balanced equation, it'll be impossible to do the next step.

00:41 From this one's going to be the hardest one.

00:43 And i like to kind of look at the most unique species here, and carbons are most unique ones.

00:51 So we have six carbons here.

00:53 So we'll go ahead and put six co2s, right? because co2 is always a source of carbon.

00:57 So we need at least 6 co2s to match a 6 carbon right here.

01:03 The next most unique one is h because h is only in the species and h is only in the species.

01:09 So let's go ahead and match up.

01:10 We have 12 here.

01:11 We have 2 here, so we need at least 6 here to make that possible.

01:15 All right, so let's add up our species.

01:17 We have 6 carbons.

01:19 We have 12 h's.

01:22 Then we have 12 oxygens plus 6, so we have 18 oxygens.

01:31 We have 6 carbons, 12 hydrogens, 6 oxygens there.

01:39 So we have 18 oxygens and we need 12, 18 minus 6.

01:46 So we need 12 oxygens over here.

01:48 So we need a 6 over here to give us our 12 oxygens to account for what we haven't used to have.

01:54 All right, so now the total of these should be 18 oxygens, 18 oxygens, 12 and 6.

02:00 All right, so this one is balanced.

02:03 All right, so this one, we have three options and we need, and we have two oxygen over here.

02:09 It's going to be a matter of simply multiplenies, or this number goes here and that number goes there, and so now we have six oxygens and six oxygens.

02:17 Check.

02:20 All right, this one, we have hydrogen and oxygen.

02:23 Again, focused on the species that isn't in, you know, two different species at once.

02:31 So hydrogens, they match, but we have two oxygens here and three oxygen.

02:36 Here if we put it to here that means we need to put two here and so now we have four hydrogens four oxygens four hydrogens two oxygens plus two oxygens all right and so this one is balanced too right so now comes apart we're writing out the rate expressions for these reactions so the rate of the reaction overall is equal to one over the coefficient.

03:10 So i'll put that in red.

03:13 I'll put a for coefficient times the change and the concentration of a or the amount of a you have over some time interval.

03:29 So this is per second.

03:31 So our time interval here goes one second.

03:34 So anything divided by one is just kind of null.

03:36 So we can kind of ignore dt at this point.

03:41 So i'll go ahead, kind of erase that just for simplicity purposes.

03:51 And rate is always positive.

03:53 So if you're losing something, you can just, there's a negative here, if it's a reactant, so it's always going to be positive.

04:00 But again, for simplicity, i'm just kind of ignore it, assume everything's going to be positive.

04:05 So what we have here is the rate of o2 formation, which is not the rate of the reaction.

04:14 The rate of the reaction is 1 over 6.