Question

The reaction is 50% complete in 2 hours and 75% complete in 4 hours. The order of the reaction is
(A) 0
(B) 1
(C) 2
(D) 3

Answer 1

Hint:Q1QWS 12W> For us to be able to solve the question, we first need to know what is the order of a reaction. To know that, we should know the rate law. It links the rate of a reaction to the concentration of the reactants and some constant parameters.

Complete step by step solution:
-The rate of law can be shown as
\[rate=k{{\left[ A \right]}^{x}}{{\left[ B \right]}^{y}}\]
where A and B are reactants of the equation and k is a constant parameter. The sum of x and y gives the overall order of the reaction.
-If the rate of the equation does not depend on the reactant concentration;i.e; it is constant irrespective of the reactant concentration, then the equation is zero order.
-If the rate depends on one reactant, it is a first-order reaction and if it depends on two reactants or square of 1 reactant, then it is a second-order reaction.
-The order can also be found by knowing the time duration in which the reaction is being completed. The duration in which half of the reaction gets completed is called the half-life of the reaction and is denoted by ${{t}_{1/2}}$.
-In the question, it is given that 2 hours are required to complete 50% of the reaction.
\[\Rightarrow \] half of the reaction is being done in 2 hours. Thus, ${{t}_{1/2}}$= 2 hrs initially.
Now, it is said that 75% reaction is being completed in 4 hours.
-Here, we need to observe that after 2 hours, half of the reaction was complete and it took 2 hours to do so. Now 50% reaction is only left.
We know 50+25=75 and 25 is half of 50. Also, 2 is half of 4.
-Thus observing this fact, we can deduce that the next half of the reaction is completed in 4-2=2 hours.
So, even for this, we get ${{t}_{1/2}}$=2 hrs.
-Hence, we can see that the half-life is constant. This is the case with first-order reactions only.

Therefore, the answer to the question is option (B)- 1

Note: The equation for first order reaction is $\ln \left[ A \right]=-kt+\ln \left[ {{A}_{0}} \right]$
At half-life, the reactant conc. becomes half.
So, after t=${{t}_{1/2}}$ , we can write,
\[{{\left[ A \right]}_{1/2}}/\left[ {{A}_{0}} \right]=1/2={{e}^{-k{{t}_{1/2}}}}\]
Taking log on both sides we get,
ln 0.5=-kt
Thus we get ${{t}_{1/2}}$= ln 2/k
Hence, half life of first order reaction is constant.