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Half life

For a reactio...

Question

For a reaction $A \to$ Product, graph of concentration of reactant versus times is given. The rate (in $M s^{- 1}$ ) of reaction after $53.34$ second from starting is:

A

0.693

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B

2 \times 10^{- 4}

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C

3 \times 10^{- 2}

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D

10^{3}

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Solution

The correct option is B $2 \times 10^{- 4}$
Rate is the change of concentration per unit time. For a straight line, of concentration vs time, rate is the slope of the graph.
Here, slope is $= - \frac{0 - 0.04}{200} = \frac{0.02}{200} = 2 \times 10^{- 4}$ M/s.

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Q. The rate of a reaction starting with initial concentration of

2 \times 10^{- 3} M

1 \times 10^{- 3} M

240 \times 10^{- 4} M s^{- 1}

0.60 \times 10^{4} M s e c^{- 1}

respectively. The rate constant is:

Q. The following data pertains to the reaction between A and B:

S. No	[A]	[B]	Rate
$1$	$1 \times 10^{- 2}$	$2 \times 10^{- 2}$	$2 \times 10^{- 4}$
$2$	$2 \times 10^{- 2}$	$2 \times 10^{- 2}$	$4 \times 10^{- 4}$
$3$	$2 \times 10^{- 2}$	$4 \times 10^{- 2}$	$8 \times 10^{- 4}$

Which of the following inferences can be drawn from the above data?

1.

Rate law of the reaction is

K [A] [B]

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2.

Rate constant of the reaction is

10^{- 4}

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3.

Rate of reaction increases four times by doubling the concentration of each reactant.

Select the correct answer from the codes given below:

Sr.No.	$[A]_{0} / M$ (lonic conc.)	$[B]_{0} / M$ (lonic conc.)	rate/ $(M s e c^{- 1})$
1	$1.6 \times 10^{- 3}$	$5 \times 10^{- 2}$	$10^{- 3}$
2	$3.2 \times 10^{- 3}$	$5 \times 10^{- 2}$	$4 \times 10^{- 3}$
3	$1.6 \times 10^{- 3}$	$10^{- 1}$	$2 \times 10^{- 3}$
4	$3.2 \times 10^{- 3}$	$10^{- 1}$	$8 \times 10^{- 3}$

For the reaction

A + B ⟶ C

; starting with different initial concentration of A and B, initial rate of reaction were determined graphically in four experiments :
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