Take Home Quiz Show all work! Aehr Due Tuesday at the start of class. For the hy

Question

Take Home Quiz Show all work! Aehr Due Tuesday at the start of class. For the hypothetical reaction 1. A+2B 2C Use the table below to determine the rate law, the overall order, value of the rate constant k, and the rate of the fourth experiment and the energy of activation. Exp. Rate(M/s Temp (C) 0.020 0.020 0.040 0.035 0.020 0.100 0.050 0.100 0.050 1.4 x 10 5.6 x 10 1.4 x 10 25.0 25.0 25.0 25.0 28.0 6.0 x 10 Explain how measuring the rate with respect to A would differ from measuring the rate with respect to B

Explanation / Answer

Solution :-

We are given with the data table for the concentration of the reactants and the rate of reaction with respect to the temperature

We need to find the order of the each reactant

To find the order with respect to A we select experiment where concentration of A is different and concentration of B is same.

Lets calculate the order with respect to reactant A using the experiment data 1 and 3

Rate 3 /rate1 = ([A]3/[A]1)m

1.4E-3/1.4E-3 = (0.04/0.020)m

1 = 2m

Log 1 = m * log 2

Log 1 / log 2 = m

0=m

So the order with respect to A is zero order

Now lets calculate order with respect to B using experiment data 1 and 2

Rate2/rate1 = ([B]2/[B]1)n

5.6E-3/1.4E-3 = (0.100 / 0.050)n

4 = 2n

Log 4 = n * log 2

Log 4/ log 2 = n

2 =n

Therefore order with respect to B is second order

Now lets write the rate law for the reaction

Rate = K[B]2

Now lets calculate the rate constant K

Lets use the concentration of the experiment 1 in the rate law formula

1.4E-3 = K[0.05]2

1.4E-3 / [0.05]2= K

0.56 M-1 s-1= K

Now lets calculate the rate for the experiment 4

Rate = K[B]2

Rate = 0.56 M-1 s-1 * [0.100]2

Rate = 5.6*10-3 M/s

Now lets calculate the activation energy using the data from the experiment 4 and 5

T1 = 25.0 C +273 = 298 K

T2 = 28.0 C +273 = 301 K

Rate 4 =0.0056

Rate 5= 0.006

We need to calculate the rate constant at 28 C

We know rate equation

Rate = k[B]2

6.0E-3 = k [ 0.050]2

6.0E-3 /[ 0.050]2 =k

2.4 = k

Now lets use the Arrhenius equation to calculate the activation energy

Formula

ln (k2/k1) = Ea / R [(1/T1)-(1/T2)]

now lets put the values in the formula (R= 8.314 J/ mol K)

ln(2.4/0.56) = Ea / 8.314 J per K mol [(1/298)-(1/301)]

1.455 = Ea / 8.314 J per K mol * 3.344E-5

1.455 * 8.314 J per K mol /3.344E-5 = Ea

361748 J per mol = Ea

Convert it to kJ per mol

361748 J per mol * 1 kJ / 1000 J =362 kJ/ mol

So activation energy is 362 kJ/ mol

Since the order of the reaction with respect A is zero order therefore if the rate is determined using the A instead of the B then it will not show any change in the rate with the concentration of A

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