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 t_ 1/8 and t_ 1/10 respectively. What is the value of t_ 1/8

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 9

For first-order decomposition with rate equation

[A] = [A]_0 e^{-kt}

, the time for concentration to reduce to fraction f is:

t = \frac{1}{k}\ln\frac{1}{f} = \frac{\ln(1/f)}{k}

For reduction to 1/8:

t_{1/8} = \frac{\ln 8}{k} = \frac{3\ln 2}{k}

For reduction to 1/10:

t_{1/10} = \frac{\ln 10}{k}

Therefore:

\frac{t_{1/8}}{t_{1/10}} = \frac{\ln 8}{\ln 10} = \frac{3\log 2}{\log 10} = \frac{3 \times 0.3}{1} = 0.9

Thus:

\frac{t_{1/8}}{t_{1/10}} \times 10 = 0.9 \times 10 = 9

An organic compound undergoes first order decomposition. The time...

Options

Correct answer

Step-by-step solution

More from Chemical Kinetics