Question:
For the reaction : $ \ce{H_2 +I_2 -> 2HI} $ , the differential rate law is
For the reaction : $ \ce{H_2 +I_2 -> 2HI} $ , the differential rate law is
Updated On: Aug 2, 2024
- $ -\frac{d\left[H_{2}\right]}{dt}=-\frac{d\left[I_{2}\right]}{dt}=2 \frac{d\left[HI\right]}{dt} $
- $ -\frac{d\left[H_{2}\right]}{dt}=-2\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt} $
- $ -\frac{d\left[H_{2}\right]}{dt}=-\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt} $
- $ -\frac{d\left[H_{2}\right]}{dt}=\frac{d\left[I_{2}\right]}{dt}=\frac{d\left[HI\right]}{dt} $
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The Correct Option is B
Solution and Explanation
$H _{2}+ I _{2} \longrightarrow 2 HI$
Rate of reaction $=\frac{-d\left[ H _{2}\right]}{d t}$
$=\frac{-d\left[ I _{2}\right]}{d t}= \frac{1}{2} \frac{d[ HI ]}{d t}$
or $\frac{-2 d\left[ H _{2}\right]}{d t}=\frac{-2 d\left[ I _{2}\right]}{d t}$
$=\frac{d[ HI ]}{d t}$
Rate of reaction $=\frac{-d\left[ H _{2}\right]}{d t}$
$=\frac{-d\left[ I _{2}\right]}{d t}= \frac{1}{2} \frac{d[ HI ]}{d t}$
or $\frac{-2 d\left[ H _{2}\right]}{d t}=\frac{-2 d\left[ I _{2}\right]}{d t}$
$=\frac{d[ HI ]}{d t}$
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Concepts Used:
Rate of a Chemical Reaction
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or products per unit time.
Consider the reaction A → B,
Rate of the reaction is given by,
Rate = −d[A]/ dt=+d[B]/ dt
Where, [A] → concentration of reactant A
[B] → concentration of product B
(-) A negative sign indicates a decrease in the concentration of A with time.
(+) A positive sign indicates an increase in the concentration of B with time.
Factors Determining the Rate of a Reaction:
There are certain factors that determine the rate of a reaction:
- Temperature
- Catalyst
- Reactant Concentration
- Chemical nature of Reactant
- Reactant Subdivision rate