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, \(k =6.0\  L^2mol^{-2} min^{-1 }\)