(a) k = \(\frac{2.303}{t}\) log \(\frac{[R]_o}{[R]}\)
For three-forth of the reaction to take place,
[R] = \(\frac{[R]_o}{[4]}\) or \(\frac{[R]_o}{[R]}\) = 4
Substituting the vale in the rate equation, we have
2.54 x 10-3 = \(\frac{2.303}{t}\)log 4
or t = \(\frac{2.303}{2.54}\) x 103 x 0.6021 = 5.46 x 102s
b) (i) dx/dt = k [A][B]2
(ii) 9 times
(iii) 8 times