Question

A first order reaction is 50% complete in ${ 25 }$ minutes. Calculate the time for 80% completion of the reaction.

Answer 1

Hint: First order reaction: A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration.
- The rate increases as the number of times the concentration of reaction is increased.

Complete Solution:
In chemistry, the law of definite proportion, sometimes called Proust's law, or law of constant composition states that a given chemical compound always contains its component elements in fixed ratio and does not depend on its source and method of preparation. For example, in a nitrogen dioxide (\[N{{O}_{2}}\]) molecule, the ratio of the number of nitrogen and oxygen atoms is$1:2$but the mass ratio is$14:32$(or$7:16$).
Now, let us look at the Haber’s process of Ammonia production and observe the reaction that takes place in the same.
The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.
${{N}_{2}}\,+\,3{{H}_{2}}\,\xrightarrow[KOH]{Fe}\,\,\,2N{{H}_{3}}$
The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency.
Now, let us apply this knowledge to the given question.
$\begin{matrix}
{{N}_{2}} \\
1\text{ volume} \\
10\text{ litres} \\
\end{matrix}\text{ }\begin{matrix}
 + \\
 {} \\
{} \\
\end{matrix}\text{ }\begin{matrix}
3{{H}_{2}} \\
\text{3 volume} \\
30\text{ litres} \\
\end{matrix}\text{ }\begin{matrix}
 \to \\
 {} \\
 {} \\
\end{matrix}\text{ }\begin{matrix}
2N{{H}_{3}} \\
\text{2 volume} \\
\text{20 litres} \\
\end{matrix}$
It is given that only $50%$ of the expected product is formed, hence only $10$ litres of \[N{{H}_{3}}\] is formed.
\[{{N}_{2}}\operatorname{used}=5\operatorname{Litres}\operatorname{left}=30-5=25\operatorname{Litres}\]
\[{{H}_{2}}\operatorname{used}=15\operatorname{Litres}\operatorname{left}=30-15=15\operatorname{Litres}\]

Note: The possibility to make a mistake is that you may confuse between log and ln in the formulas. ${ ln = 2.303~log }$ Don’t forget to apply this formula. Also, the first-order reaction at half-life is decreased by half of its original value.