1. Calculate the rate constant, k, and write the rate law...

1. Calculate the rate constant, k, and write the rate law expression for the catalyzed decomposition of hydrogen peroxide. Explain how you determined the order of the reaction in H2O2 and KI. Rate 1 / Rate 2 = (1.34 * 10^-4) / (8.12 * 10^-5) = (k[H2O2]^x [I-]^y) / (k[H2O2]^x [I-]^y); 1.65 = (k[0.706]^x [0.1]^y) / (k[0.706]^x [0.05]^y); 1.65 = 2^y; y = 0.723 Rate 1 / Rate 3 = (1.34 * 10^-4) / (6.93 * 10^-5) = 1.93 = (k[0.706]^x [0.1]^y) / (k[0.353]^x [0.1]^y); 1.93 = 2^x; x = 0.951; Rate = k[H2O2]^1[I-]^1 k = (1.34 * 10^-4) / (0.706 * 0.1); k = 1.89 * 10^-3 2. The following mechanism has been proposed for this reaction: H2O2 + I- ? IO- + H2O H2O2 + IO- ? I- + H2O + O2 If this mechanism is correct, which step must be the rate-determining step? Explain. Step 1, because both coefficients are one the order of the proposed law would match our determined experimental rate law with the 1st step being the rate-determining step. 3. Use the Arrhenius equation (shown below) to determine the activation energy, Ea, for this reaction. R = 8.314 J/mol · K ln(k1/k2) = (Ea/R)((1/T2) - (1/T1))

Transcript

00:01 Hello students, we are supposed to find out rate of reaction in the question and give us clear explanation for this.

00:14 The keynote here is we have to mention the steps involved in calculation of rate of reaction and the rate constants.

00:33 Let us start the solution here.

00:36 Initial rate of reaction this is given in the question for h2 .02 and minus let me start with 0 .0 811 next we have minus 0 .0182 0 .0 .612 minus 0 .0 .612 minus 0 .02587.

01:09 Let us see for h2 initial rate 0 .704 so for h2o2 after mixing it is 0 .0070 -24.

01:24 Next, it is 0 .70 -4.

01:30 Whereas in the third, after mixing, we have 0 .352.

01:38 Last, these are the moles for h2.

01:45 Next, i minus 0 .1 mole, 0 .05 mole.

01:54 0 .1 mole, 0 .05 moles.

02:00 Now let us move to the rate of the reaction.

02:07 Now, let us observe.

02:10 Rate is equal to, that is rate of reaction is equal to k multiplied by h2o2 to the power x, multiplied by i minus to the power y.

02:23 Now let us take 1 2 3 4 be rate 1 rate 2 and rate 3 and rate 4 so these are the rate reactions now let us start the calculation rate 1 divided by rate 4 is equal rate 1 divided by rate 2 is equal to k multiplied by h2 o 2 to the power x 1 1 divided by k h2 o2 to the power x to i minus to the power y 2 now rate of reaction is always positive the rate of reaction is always positive that is 0 .0 8 double 1 divided by this is supposed to k multiplied by 0 .0 .0 .8 1 divided by this is supposed to k multiplied by 0 .0 .0.

03:59 704 to the power x multiplied by 0 .1 to the power y divided by k multiplied by 0 .05 to the power y.

04:19 Now simplify the equation cancel both the numerator and denominator 0 .704 to the power x simplify.

04:29 Now take logarithm on both sides.

04:34 Log on both sides after the resting log log log 4 .456 is equals to y log 2 0 .648 equal to y multiplied by 0 .3010.

05:02 Let us calculate 5 .648 divided by 0 .3013 .3010 that is equal to 2 .3010 that is equal to 2 .10.

05:13 This is equal to 2 .15 that is y is more or less equal to now let's so order with respect to i minus is two now order is equal to rate 1 divided by rate 3 this is equal to k multiplied by h2 o 2 minus to the power y k multiplied by h2 3 i minus 3 here to the power y now rest to the values 0 .08 3 is 0 .0612 which is equal to k multiplied by 0 .704 to the power x multiplied by 0 .704 to the power x multiplied by 0 .1 to the power y divided by k 5 to the power x multiplied by 0 .1 to the power y.

06:53 1 .325 is equal to 2 to the power x.

07:02 Now take log on both sides...

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