1) For the first order reaction:

k = (2.303 / t) log (a / a – x)

When x = (40 / 100) a = 0.4 a

t = 50 minutes (given)

Therefore, k = (2.303 / 50) log (a / a – 0.4 a)

k = (2.303 / 50) log (1 / 0.6)

= 0.010216 min^{-1}

Hence the value of the rate constant is 0.010216 min^{-1}

2) t= ?, when x = 0.8 a

From above, k = 0.010216 min^{-1}

Therefore, t = (2.303 / 0.010216) log (a / a – 0.8 a)

= (2.303 / 0.010216) log (1 / 0.2) = 157.58 min

The time at which the reaction will be 80% complete is 157.58 min.