Hydrolysis of benzene diazonium chloride follows first order kinetics. The time taken for its decomposition to 1/8 and 1/10 of its original concentration are $[t_{1/8}]$ and $[t_{1/10}]$ respectively. What is the ratio of $[t_{1/8}]$ to $[t_{1/10}]$? ($\log 2=0.30$, $\log 3=0.48$, $\log 4=0.60$)

Question

For a reaction $ A + B \to \text{products} $, the rate of the reaction was doubled when the concentration of $ A $ was doubled. When the concentrations of $ A $ and $ B $ were doubled, the rate was again doubled. The order of the reaction with respect to $ A $ and $ B $ are:

Answer 1

Step 1: Recall the first order time formula.
For a first order reaction, the time to drop to a fraction $1/n$ of the start is $t = \dfrac{2.303}{k}\log n$. We will use this for both fractions and then take the ratio, so $k$ cancels nicely.
Step 2: Write the time for the 1/8 decomposition.
Here the concentration falls to $1/8$, so $n = 8$. \[ t_{1/8} = \frac{2.303}{k}\log 8 \]
Step 3: Simplify log 8.
Since $8 = 2^3$, $\log 8 = 3\log 2 = 3(0.30) = 0.90$. So $t_{1/8} = \dfrac{2.303}{k}(0.90)$.
Step 4: Write the time for the 1/10 decomposition.
Here $n = 10$, and $\log 10 = 1$. \[ t_{1/10} = \frac{2.303}{k}\log 10 = \frac{2.303}{k}(1.0) \]
Step 5: Take the ratio.
Dividing the two, the $\dfrac{2.303}{k}$ factors cancel: \[ \frac{t_{1/8}}{t_{1/10}} = \frac{0.90}{1.0} = \frac{9}{10} \]
Step 6: State the answer.
So the ratio is $9:10$.
\[ \boxed{9:10} \]

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The Correct Option is A

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