Question

A first order reaction takes $23.1min$. for $50\%$ completion. Calculate the time required for $75$% completion of this reaction.$[Given:log2=0.301,log3=0.44771,\log 4=0.6021]$.

Answer 1

$\frac{t_1}{2}=23.1min$

$\frac{t_1}{2}=\frac{0.693}{k}$

k=$\frac{0.693}{23.1}$

k=0.03

k=$\frac{2.303}{t}log\frac{\left[R_0\right]}{\left[R\right]}$

0.03=$\frac{2.303}{t}log\left(\frac{100}{25}\right)$

0.03=$\frac{2.303}{t}log\left(4\right)$

t=$\frac{2.303}{0.03}log\left(4\right)$

t=46.22 min.