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 $[A] / m o l L^{- 1}$ $B] / m o l L^{- 1}$ Initial rate/ $m o l L^{- 1} m i n^{- 1}$
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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Solution

The rate law expression is:
$R a t e = k [A] [B]^{0}$

$R a t e = k [A]$

$2.0 \times 10^{- 2} = k [0.1]$

$k = 0.2 m i n^{- 1}$

From second experiment,
$4.0 \times 10^{- 2} = 0.2 [A]$

$[A] = 0.2 m o l L^{- 1}$

From third experiment,
$R a t e = 0.2 \times 0.4 = 8.0 \times 10^{- 2} m o l L^{- 1} m i n^{- 1}$

From fourth experiment,
$2.0 \times 10^{- 2} = 0.2 [A]$

$[A] = 0.1 m o l L^{- 1}$

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Similar questions

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

Q. Fill in the blanks in the following table which treats reaction of a compound

A

with a compound

B

, that is of the first-order with respect to

A

and zero-order with respect to

B

Expt. No.	$[A]$ $(m o l L^{- 1})$	$[B]$ $(m o l L^{- 1})$	Initial rate $(m o l L^{- 1} s^{- 1})$
$1.$	$0.1$	$0.1$	$2.0 \times 10^{- 2}$
$2.$	___	$0.2$	$4.0 \times 10^{- 2}$
$3.$	$0.4$	$0.4$	___
$4.$	___	$0.2$	$2.0 \times 10^{- 2}$

Q. 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	$[A] / M$	$[B] / M$	Initial rate/ $M m i n$
I	$0.1$	$0.1$	$2.0 \times 10^{- 2}$
II	....	$0.2$	$4.0 \times 10^{- 2}$
III	$0.4$	$0.4$	....
IV	....	$0.2$	$2.0 \times 10^{- 2}$

Q. Find the values of A, B and C in the following table for the reaction :

X + Y ⟶ Z

The reaction is first order with respect to

X

and zero order with respect to

Y

$E x p$	$[X] (m o l L^{- 1})$	$[Y] (m o l L^{- 1})$	$I n i t i a l r a t e (m o l L^{- 1} s - 1)$
1.	$0.1$	$0.1$	$2 \times 10^{- 2}$
2.	$A$	$0.2$	$4 \times 10^{- 2}$
3.	$0.4$	$0.4$	$B$
4.	$C$	$0.2$	$2 \times 10^{- 2}$

The reation 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:

$\begin{matrix} E x p e r i m e n t & [A] m o l L^{- 1} & [B] m o l L^{- 1} & I n i t i a l r a t e m o l L^{- 1} m i n^{- 1} I & 0.1 & 0.1 & 2.0 \times 10^{- 2} I I & - & 0.2 & 4.0 \times 10^{- 2} I I I & 0.4 & 0.4 & - I V & - & 0.2 & 2.0 \times 10^{- 2} \end{matrix}$

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

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]/molL−1B]/molL−1Initial rate/ molL−1min−1I0.10.12.0×10−2II-0.24.0×10−2III0.40.4-IV-0.22.0×10−2

The rate law expression is: Rate=k[A][B]0Rate=k[A]2.0×10−2=k[0.1]k=0.2min−1 From second experiment,4.0×10−2=0.2[A][A]=0.2 molL−1From third experiment,Rate=0.2×0.4=8.0×10−2molL−1min−1From fourth experiment,2.0×10−2=0.2[A][A]=0.1molL−1