Question

The following results have been obtained during the kinetic studies of the reaction:
$2A+B→C+D$

Experiment	A/mol L -1	B/mol L -1	Initial rate of formation of D/mol L -1min-1
I	0.1	0.1	6.0 × 10-3
II	0.3	0.2	7.2 × 10-2
III	0.3	0.4	2.88 × 10-1
Iv	0.4	0.1	2.40 × 10-2

Determine the rate law and the rate constant for the reaction.

Answer 1

Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
Rate = k [A]x[B]y
According to the question,
6.0×10-3 = k[0.1]x[0.1]y ......(i)
7.2×10-2 = k[0.3]x [0.2]y ......(ii)
2.88×10-1 = k[0.3]x[0.4]y ......(iii)
2.40×10-2 = k[0.4]x[0.1]y .......(iv)
Dividing equation (iv) by (i), we obtain

$\frac {2.40×10^{-2}}{6.0×10^{-3}}$ = $\frac {k[0.4]^x[0.1]^y }{ k[0.1]^x[0.1]^y}$

⇒ $4= \frac {[0.4]^x}{[0.1]^x}$

⇒ $4 = (\frac {0.4}{0.1})^x$

⇒ $4^1 = 4^x$

⇒ $x = 1$

Dividing equation (iii) by (ii), we obtain

Therefore, the rate law is

$Rate = k[A][B]^2$

$k = \frac {Rate}{[A][B]^2}$

From experiment I, we obtain

$k =\frac { 6.0 \times 10^{-3} mol L^{-1} min^{-1}}{(0.1\ mol L^{-1})(0.1 \ mol L^{-1})^2}$

= $6.0\ L^2mol^{-2} min^{-1 }$

From experiment II, we obtain

k = $\frac {7.2\times 10^{-2} mol L^{-1} min^{-1}}{(0.3\ mol L^{-1})(0.2\ mol L^{-1})^2}$

= $6.0 \ L^2 mol^{-2} min^{-1}$

From experiment III, we obtain

k = $\frac {2.88\times 10^{-1} mol L^{-1} min^{-1}}{(0.3\ mol L^{-1})(0.4\ mol L^{-1})^2}$

= $6.0\ L^2mol^{-2} min^{-1 }$

From experiment IV, we obtain

k = $\frac {2.40 \times 10^{-2} mol L^{-1} min^{-1}}{(0.4\ mol L^{-1})(0.1\ mol L^{-1})^2}$

= $6.0\ L^2mol^{-2} min^{-1 }$

Therefore, rate constant, $k =6.0\ L^2mol^{-2} min^{-1 }$