The rate constant for the first-order decomposition of...

The rate constant for the first-order decomposition of N2O5(g) to NO2(g) and O2(g) is 7.48 * 10-3 s-1 at a given temperature. a. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.100 atm to rise to 0.145 atm. b. Find the length of time required for the total pressure in a system containing N2O5 at an initial pressure of 0.100 atm to rise to 0.200 atm. c. Find the total pressure after 100 s of reaction.

The rate constant for the first-order decomposition of N2O5(g) to
NO2(g) and O2(g) is 7.48 * 10-3 s-1 at a given temperature.
a. Find the length of time required for the total pressure in a
system containing N2O5 at an initial pressure of 0.100 atm to
rise to 0.145 atm.
b. Find the length of time required for the total pressure in a
system containing N2O5 at an initial pressure of 0.100 atm to
rise to 0.200 atm.
c. Find the total pressure after 100 s of reaction.

Transcript

00:01 So there's a lot going on here.

00:02 Let me see if i can put it all together.

00:04 We have this chemical reaction, and they didn't give it to us balanced.

00:10 So they wanted it even more difficult than it already was.

00:14 To balance it, we have two nitrogen.

00:16 So we'll put a two here, so we have two nitrogen.

00:19 That then gives us five oxygens here.

00:23 Four plus two gives us six, so we need a one -half in front of the oxygen.

00:27 But we don't want fractions, so we'll multiply everything through by two.

00:31 And this will be our balanced chemical reaction.

00:34 Initially, we have 0 .100atm, n205, and an initial amount of zero on the other two.

00:48 Then as equilibrium, not equilibrium, but then as the reaction proceeds for a particular amount of time, which will be determining, the n205 will decrease by 2x, because of its coefficient of 2.

01:08 And then this will increase by 4x, and o2 will increase by x.

01:13 So at equilibrium, or maybe we could just say final, after a particular amount of time, we get 0 .10 minus 2x, and o2 is 4x, and o2 is x.

01:29 They tell us that the sum, after a particular amount of time, is 0 .145 atm.

01:35 This then allows us to solve for x, where the total pressure is .145 atm and is equal to .1 plus 3x.

01:49 So x then becomes equal to .015 atm, but the pressure of n205 will be .1 minus 2x at that time, or .07 atm.

02:09 Now finally, we can use the first order integrated rate law in order to calculate the time that has elapsed.

02:16 The first order integrated rate law tells us that the natural log of the amount at time t, that is .07 atm for n205, divided by the amount or pressure of n205 at time 0 .1 at time 0 .1 will be equal to negative k.

02:36 K is provided at 7 .48 times 0 .5 times 0 .1 .m...

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