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Zero Order Reaction

In a reaction...

Question

In a reaction, reactant $A$ decomposes $10$ % in $1 h o u r$ , $20$ % on $2 h o u r$ and $30$ % in $3 h o u r$ . The unit of rate constant of this reaction is:

A

s e c^{- 1}

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B

m o l l i t r e^{- 1} s e c^{- 1}

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C

l i t r e m o l^{- 1} s e c^{- 1}

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D

l i t r e^{2} m o l^{- 2} s e c^{- 1}

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Solution

The correct option is D $m o l l i t r e^{- 1} s e c^{- 1}$
If we observe the decomposition of reactant A with time, we conclude that the decomposition increases linearly with time. Hence, the reaction can be considered to be of zero order and the decomposition at any time $t$ is given by,

$\Rightarrow x = k t$
$k = \frac{x}{t}$

Therefore, unit of $k = m o l s e c^{- 1} l i t r e^{- 1}$

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Q. In a certain reaction, 10

%

of the reactant decomposes in 1 hr, 20

%

in two hours, 30

%

in three hours and so on the dimensions of the rate constant is

Q. The rate constant (K) for the reaction,

2 A + B \to

Product was found to be

2.5 \times 10^{- 5}

m o l^{- 1} s e c^{- 1}

2.60 \times 10^{- 5} l i t r e m o l^{- 1} s e c^{- 1}

after 30 sec and

2.55 \times 10^{- 5} l i t r e m o l^{- 1} s e c^{- 1}

after 50 sec. The order of reaction is:

Q. The rate constant of the reaction :

2 A + B \to C

2.57 \times 10^{- 5}

l t . m o l^{- 1}

s e c^{- 1}

2.65 \times 10^{- 5}

l t . m o l^{- 1} s e c^{- 1}

after 20 sec and

2.55 \times 10^{- 5}

l t . m o l^{- 1} s e c^{- 1}

after 30 sec. The order of the reaction is :

Q. The rate constant

(k)

for the reaction

2 A + B \to

product, was found to be

2.5 \times 10^{- 5}

{m o l}^{- 1} {s e c}^{- 1}

15

2.60 \times 10^{- 5}

{m o l}^{- 1} {s e c}^{- 1}

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2.55 \times 10^{- 5}

{m o l}^{- 1} {s e c}^{- 1}

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The order of the reaction is:

Q. The rate constant

(K)

for the reaction,

2 A + B \to

Product was found to be

2.5 \times 10^{- 5}

{m o l}^{- 1} {s e c}^{- 1}

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2.60 \times 10^{- 5}

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2.55 \times 10^{- 5}

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sec.
The order of reaction is:

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