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Standard XII

Chemistry

Zero Order Reaction

In a reaction...

Question

In a reaction; $2 A \to P r o d u c t s$ , the concentration of $A$ decreases from $0.5 m o l l i t r e^{- 1}$ to $0.4 m o l l i t r e^{- 1}$ in $10$ minutes. Calculate rate during this interval.

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Solution

$R a t e = \frac{1}{2} \frac{[A_{0}] - [A]}{t}$

$R a t e = \frac{1}{2} \frac{[A_{0}] - [A]}{t}$

$R a t e = \frac{1}{2} \frac{[0.5] - [0.4]}{10}$

$R a t e = 5 \times 10^{- 3}$ $m o l L^{- 1} {m i n}^{- 1}$

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In a reaction; 2A→Products, the concentration of A decreases from 0.5 mol litre−1 to 0.4 mol litre−1 in 10 minutes. Calculate rate during this interval.

Rate=12[A0]−[A]tRate=12[A0]−[A]tRate=12[0.5]−[0.4]10Rate=5×10−3 mol L−1 min−1

In a reaction; $2 A \to P r o d u c t s$ , the concentration of $A$ decreases from $0.5 m o l l i t r e^{- 1}$ to $0.4 m o l l i t r e^{- 1}$ in $10$ minutes. Calculate rate during this interval.

$R a t e = \frac{1}{2} \frac{[A_{0}] - [A]}{t}$

$R a t e = \frac{1}{2} \frac{[A_{0}] - [A]}{t}$

$R a t e = \frac{1}{2} \frac{[0.5] - [0.4]}{10}$

$R a t e = 5 \times 10^{- 3}$ $m o l L^{- 1} {m i n}^{- 1}$