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Question: Fill in the blanks in the following table for the reaction \(X + Y \rightarrow Z\) the reaction is o...

Fill in the blanks in the following table for the reaction X+Y→ZX + Y \rightarrow Z the reaction is of first order w.r.t X and zero order w.r.t. Y.

Exp.[X](molL−1)[Y](molL−1)

Initial rate

(molL−1s−1)

1.0.10.12 × 10−2
2.A0.24 × 10−2
3.0.40.4B
4.C0.22 × 10−2
A

A=0.2molL−1,B=8×10−2molL−1s−1,C=0.1molL−1A = 0.2molL^{- 1},B = 8 \times 10^{- 2}molL^{- 1}s^{- 1},C = 0.1molL^{- 1}

B

A=0.4molL−1,B=4×10−2molL−1s−1,C=0.2molL−1A = 0.4molL^{- 1},B = 4 \times 10^{- 2}molL^{- 1}s^{- 1},C = 0.2molL^{- 1}

C

A=0.2molL−1,B=2×10−2molL−1s−1,C=0.4molL−1A = 0.2molL^{- 1},B = 2 \times 10^{- 2}molL^{- 1}s^{- 1},C = 0.4molL^{- 1}

D

A=0.4molL−1,B=2×10−2molL−1s−1,C=0.4molL−1A = 0.4molL^{- 1},B = 2 \times 10^{- 2}molL^{- 1}s^{- 1},C = 0.4molL^{- 1}

Answer

A=0.2molL−1,B=8×10−2molL−1s−1,C=0.1molL−1A = 0.2molL^{- 1},B = 8 \times 10^{- 2}molL^{- 1}s^{- 1},C = 0.1molL^{- 1}

Explanation

Solution

Rate=k[X][Y]ºRate = k\lbrack X\rbrack\lbrack Y\rbrack º

Rate is independent of conc. of Y and it depends only on the conc. of X and it is the first order reaction.

From exp. (1), 2×10−2=k(0.1)2 \times 10^{- 2} = k(0.1) ….. (i)

Form exp. (2), 4×10−2=k(A)4 \times 10^{- 2} = k(A) ….. (ii)

Dividing (ii) and (i), 4×10−22×10−2=k(A)k(0.1)=A0.1\frac{4 \times 10^{- 2}}{2 \times 10^{- 2}} = \frac{k(A)}{k(0.1)} = \frac{A}{0.1}

⇒2×0.1=A\Rightarrow 2 \times 0.1 = A

⇒A=0.2molL−1\Rightarrow A = 0.2molL^{- 1}

From exp. (3), B=k(0.4)B = k(0.4) … (iii)

Diving (iii) and (i), B2×10−2=k(0.4)k(0.1)=4\frac{B}{2 \times 10^{- 2}} = \frac{k(0.4)}{k(0.1)} = 4

⇒B=4×2×10−2=8×10−2molL−1s−1\Rightarrow B = 4 \times 2 \times 10^{- 2} = 8 \times 10^{- 2}molL^{- 1}s^{- 1}

From exp. (4), 2×10−2=k(C)2 \times 10^{- 2} = k(C) … (iv)

Dividing (iv) and (i), 2×10−22×10−2=k(C)k(0.1)=C0.1\frac{2 \times 10^{- 2}}{2 \times 10^{- 2}} = \frac{k(C)}{k(0.1)} = \frac{C}{0.1}

⇒C=0.1molL−1\Rightarrow C = 0.1molL^{- 1}