Question:
For the reaction $N_{2}+3H_{2} \rightarrow 2NH_{3}$
if $\frac{\Delta\left[NH_{3}\right]}{\Delta t}=2\times10^{-4}\,mol\,l^{-1}s^{-1},$ then value of $\frac{-\Delta\left[H_{2}\right]}{\Delta t}$ would be
For the reaction $N_{2}+3H_{2} \rightarrow 2NH_{3}$
if $\frac{\Delta\left[NH_{3}\right]}{\Delta t}=2\times10^{-4}\,mol\,l^{-1}s^{-1},$ then value of $\frac{-\Delta\left[H_{2}\right]}{\Delta t}$ would be
Updated On: Mar 30, 2024
- $1 � 10^{-4}\, mol\, L^{-1}s^{-1}$
- $3 � 10^{-4}\, mol\, L^{-1}s^{-1}$
- $4 � 10^{-4}\, mol\, L^{-1}s^{-1}$
- $6 � 10^{-4}\, mol\, L^{-1}s^{-1}$
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The Correct Option is B
Solution and Explanation
$N_{2}+3H_{2} \rightarrow 2NH_{3}$
$\frac{-\left[N_{2}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[H_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[NH_{3}\right]}{\Delta t}$
$\therefore \frac{-\Delta\left[H_{2}\right]}{\Delta t}=\frac{3}{2}\times\frac{\Delta\left[NH_{3}\right]}{\Delta t}=\frac{3}{2}\times2\times10^{-4}$
$=3\times10^{-4}\,mol\,L^{-1}S^{-1}$
$\frac{-\left[N_{2}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[H_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[NH_{3}\right]}{\Delta t}$
$\therefore \frac{-\Delta\left[H_{2}\right]}{\Delta t}=\frac{3}{2}\times\frac{\Delta\left[NH_{3}\right]}{\Delta t}=\frac{3}{2}\times2\times10^{-4}$
$=3\times10^{-4}\,mol\,L^{-1}S^{-1}$
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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