Question

A graph plotted between $log t_{50}$ vs log concentration is a straight line. What conclusion can you draw from the given graph?



[Given: n = order of reaction]

• A. $\text{n} \, \text{=} \, \text{1,} \, \text{t}_{\frac{\text{1}}{\text{2}}} \text{=} \frac{\text{1}}{\text{k} \text{.a}}$
• B. $n=2,t_{\frac{1}{2}}=\frac{1}{a}$
• C. $\text{n} \, \text{=} \, \text{1,} \, \text{t}_{\frac{\text{1}}{\text{2}}} \, \text{=} \, \frac{\text{0} \text{.693}}{\text{k}}$
• D. None of the above

Answer 1

Answer: (C)

$t_{\frac{1}{2}} \propto \left(a\right)^{1 - n}$

or $t_{\frac{1}{2}}=Z\left(a\right)^{^{1 - n}}$

or $log t_{\frac{1}{2}}=log ⁡ Z+\left(\right.1-n\left.\right)log ⁡ a$

or $\left(\right.y\left.\right)=c+mx$

Thus, slope = m = 1 - n

or 1 - n = 0

So n = 1

and for I order reaction $t_{\frac{1}{2}}=\frac{0.693}{k}$ .