(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 10^{3} x 0.6021 = 5.46 x 10^{2}s

b) (i) dx/dt = k [A][B]^{2}

(ii) 9 times

(iii) 8 times