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Chemistry Question on Chemical Kinetics

During the kinetic study of the reaction, 2A+B⟶C+D2A + B\longrightarrow C + D, following results were obtained . Based on the above data which one of the following is correct? RunA/mol L−1A/mol\,L^{-1}B/mol L−1B/mol\,L^{-1}Initial rate of formation of D/mol L−1 min−1D/mol\,L^{-1}\,min^{-1}
I0.10.16.0×10−36.0 \times 10^{-3}
II0.30.27.2×10−27.2 \times 10^{-2}
III0.30.42.88×10−12.88 \times 10^{-1}
IV0.40.12.40×10−22.40 \times 10^{-2}
A

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

B

Rate = k[A][B]

C

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

D

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

Answer

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

Explanation

Solution

Let the order of reaction with respect to A is x
and with respect to B is y. Thus,
rate=k[A]x[B]y[A]^x [B]^y
(xx and yy are stoichiometric coefficient)
For the given cases,
I. rate = k(0.1)x (0.1)y=6.0×10−3k (0.1)^x \, (0.1)^y =6.0 \times 10^{-3}
II. rate = k(0.3)x (0.2)y=7.2×10−2k (0.3)^x \, (0.2)^y =7.2 \times 10^{-2}
III. rate = k(0.3)x (0.40)y=2.88×10−1k (0.3)^x \, (0.40)^y =2.88 \times 10^{-1}
IV. rate = k(0.34)x (0.1)y=2.40×10−2k (0.34)^x \, (0.1)^y =2.40 \times 10^{-2}
Dividing E (I) by E (IV), we get
(0.10.4)x(0.10.1)y=6.0×10−32.4×10−2\big(\frac{0.1}{0.4}\big)^x \big(\frac{0.1}{0.1}\big)^y =\frac{6.0 \times 10^{-3}}{2.4 \times 10^{-2}}
or (14)x=(14)1 \big(\frac{1}{4} \big)^x = \big(\frac{1}{4} \big)^1
\therefore \hspace25mm x=1
On dividing E (II) by E (Ill), we get
(0.30.3)x(0.20.4)y=7.2×10−22.88×10−1\big(\frac{0.3}{0.3}\big)^x \big(\frac{0.2}{0.4}\big)^y =\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}
or (12)y=14\big(\frac{1}{2} \big)^y = \frac{1}{4}
or (12)y=(12)2 \big(\frac{1}{2} \big)^y = \big(\frac{1}{2} \big)^2
\therefore \hspace25mm y=2
Thus, rate law is,
rate=k[A]1[B]2rate = k[A]^1 [B]^2
=k[A][B]2= k[A][B]^2