Question

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration at ${\mathrm{t}}_{1/8}$and ${\mathrm{t}}_{1/10}$ respectively. What is the value of $\frac{{\mathrm{t}}_{1/8}}{{\mathrm{t}}_{1/10}}\times 10$?

Answer 1

First-order decomposition:

• A first-order reaction is one in which the rate of reaction is strictly dependent on the concentration of only one ingredient.
• In other terms, a first-order reaction is a chemical reaction in which the rate fluctuates as the concentration of only one of the reactants changes.

Step 1 : Analyzing given quantities

• $\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)$
• where a = initial concentration
• and x = change in concentration

Step 2: finding ${\mathrm{t}}_{1/8}$and ${\mathrm{t}}_{1/10}$

• t when decomposition is $1/8$ of its initial concentration
• $\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \left(\frac{1}{1/8}\right)=\frac{2.08}{\mathrm{k}}$
• t when decomposition is $1/10$ of its initial concentration
• $\mathrm{t}=\frac{2.303}{\mathrm{k}}\log \left(\frac{1}{1/10}\right)=\frac{2.303}{\mathrm{k}}$

Step 3 : Finding $\frac{{\mathrm{t}}_{1/8}}{{\mathrm{t}}_{1/10}}\times 10$

• $\frac{{\mathrm{t}}_{1/8}}{{\mathrm{t}}_{1/10}}\times 10=\frac{{\displaystyle \frac{2.08}{\mathrm{k}}}}{{\displaystyle \frac{2.303}{\mathrm{k}}}}\times 10=9$
• So $\frac{{\mathrm{t}}_{1/8}}{{\mathrm{t}}_{1/10}}\times 10$ is $9$