question_answer

Plotting a graph of log \[{{t}_{1/2}}\] against \[log{{\left[ A \right]}_{0}}\] of a reactant for a first order reaction, the slope will be

A) \[-1\]                

B)        \[-2\]   

C) 0         

D)        +1

Correct Answer: C

Solution :

[c] \[{{t}_{1/2}}=\frac{0.693}{k}\] \[log\text{ }{{t}_{1}}/2=log0.693-logk=constant\]