Question

For the reaction
A + B + C → D
The following data were obtained at constant temperature:

Trial	Initial [A] (mol/L)	Initial [B] (mol/L)	Initial [C] (mol/L)	Initial Rate (mol/L . s)
1	0.2790	0.1670	0.0334	1.190
2	0.5580	0.1670	0.0334	2.381
3	0.5580	0.3340	0.0334	9.524
4	0.2790	0.1670	0.0668	1.190

1. What is the order with respect to each reactant? (Enter the order for A as Answer 1, that for B as Answer 2, and that for C as Answer 3. You must get all 3 correct.)

2. What is the value of the rate constant for the reaction at this temperature?

Answer 1

Answer –

We are given, reaction – A + B + C -------> D

We assume rate law is ,Rate = k [A]^x [B]^y [C]^z

The x,y and z are the order with respect to A, B and C

So,

Rate1 = k [A]₁^x [B]₁^y [C]₁^z

Rate2 = k [A]₂^x [B]₂^y [C]₂^z

Rate3 = k [A]₃^x [B]₃^y [C]₃^z

Rate4 = k [A]₄^x [B]₄^y [C]₄^z

Now we need to calculate order with respect to A means x, so we need to take ration Rate 2 by Rate 1

Rate2/ Rate1 = k [A]₂^x [B]₂^y [C]₂^z / k [A]₁^x [B]₁^y [C]₁^z

2.381 / 1.190 = (0.5580)^x /(0.2790)^x * (0.1670)^y /(0.1670)^y *(0.0334)^z /(0.0334)^z

   2 = (2)^x

So, x = 1

Now we need to calculate z

Rate3/ Rat21 = k [A]₃^x [B]₃^y [C]₃^z / k [A]₂^x [B]₂^y [C]₂^z

9.524 / 2.381 = (0.5580)^x /(0.5580)^x * (0.3340)^y /(0.1670)^y *(0.0334)^z /(0.0334)^z

    4 = (2)^y

So, y = 2

Now we need z calculate

Rate4/ Rate1 = k [A]₄^x [B]₄^y [C]₄^z / k [A]₁^x [B]₁^y [C]₁^z

1.190 / 1.190 = (0.2790) / (0.2790) * (0.1670)² /(0.1670)² *(0.0668)^z /(0.0334)^z

1 = (2)^z

So, z = 0

So the order with respect to A is 1, for B it is 2 and for C it is 0. So rate of reaction depends on A and B.

So, rate law

Rate = k [A] [B]²

Now we need to put the values and calculate k

1.190 Ms^-1 = k (0.2790)*(0.1670)²

k = 1.190 /0.00778

k = 153 M^-2 s^-1