The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment [A]/mol L^ -1 [B]/mol L^ -1 Initial rate/mol L^ -1 min^ -1 I 0.1 0.1 2.0 10^ -2 II - 0.2 4.0 10^ -2 III 0.4 0.4 - IV - 0.2 2.0 10^ -2

The given reaction is of the first order with respect to A and of zero order with respect to B. Therefore, the rate of the reaction is given by, Rate = k[A]^1 [B]^0 Rate = k[A] From experiment I, we obtain 2.0 10^ -2 mol L^ -1 ^ -1 = k(0.1 mol L^ -1 ) k = 0.2 ^ -1 From experiment II, we obtain 4.0 10^ -2 mol L^ -1 ^ -1 = 0.2 ^ -1 [A] [A] = 0.2 mol L^ -1 From experiment III, we obtain Rate = 0.2 ^ -1 0.4 mol L^ -1 = 0.08 mol L^ -1 min^ -1 From experiment IV, we obtain 2.0 10^ -2 mol L^ -1 ^ -1 = 0.2 ^ -1 [A] [A] = 0.1 mol L^ -1

The reaction between A and B is first order with respect to A and…

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The reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:

Experiment	$[A]/mol\ L^{-1}$	$[B]/mol\ L^{-1}$	Initial rate$/mol\ L^{-1}\ min^{-1}$
$I$	$0.1$	$0.1$	$2.0 \times 10^{-2}$
$II$	$-$	$0.2$	$4.0 \times 10^{-2}$
$III$	$0.4$	$0.4$	$-$
$IV$	$-$	$0.2$	$2.0 \times 10^{-2}$

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Answer

The given reaction is of the first order with respect to $A$ and of zero order with respect to $B.$
Therefore, the rate of the reaction is given by,
Rate $= k[A]^1 [B]^0$
Rate $= k[A]$
From experiment I, we obtain
$2.0 \times 10^{-2} \ mol\ L^{-1} \ \min^{-1} = k(0.1\ \ mol\ L^{-1})$
$k = 0.2\ \min^{-1}$
From experiment II, we obtain
$4.0 \times 10^{-2} \ mol\ L^{-1} \min^{-1} = 0.2 \min^{-1} [A]$
$[A] = 0.2 \ mol\ L^{-1}$
From experiment III, we obtain
Rate $= 0.2 \min^{-1} \times 0.4 \ mol\ L^{-1}$
$= 0.08 \ mol\ L^{-1}\ min^{-1}$
From experiment IV, we obtain
$2.0 \times 10^{-2} \ mol\ L^{-1}\ \min^{-1} = 0.2 \min^{-1}[A]$
$[A] = 0.1 \ mol\ L^{-1}$

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