An organic compound undergoes first order decomposition. The time taken for decomposition to (1/8)th and (1/10)th of its initial concentration are t1/8 and t1/10 respectively. What is the value of (t1/8 / t1/10) × 10
Q. An organic compound undergoes first order decomposition. The time taken for decomposition to (1/8)th and (1/10)th of its initial concentration are t1/8 and t1/10 respectively.

What is the value of (t1/8 / t1/10) × 10 ?

(log 2 = 0.3)

(A) 30

(B) 9

(C) 3

(D) 0.9

Correct Answer: 9

Explanation

For a first order reaction, the time required for concentration to reduce from initial value a to a fraction a/x is given by:

t = (2.303 / k) log x

For decomposition to (1/8)th of initial concentration:

t1/8 = (2.303 / k) log 8

Since:

log 8 = log (2³) = 3 log 2 = 3 × 0.3 = 0.9

So:

t1/8 = (2.303 / k) × 0.9

For decomposition to (1/10)th of initial concentration:

t1/10 = (2.303 / k) log 10 = (2.303 / k) × 1

Now:

t1/8 / t1/10 = 0.9

Therefore:

(t1/8 / t1/10) × 10 = 0.9 × 10 = 9

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