The half-life period of a ( 1^ s t ) order reaction is ( 60 ) minutes. What percentage will be left over after ( 240 ) minutes?

Step-by-step solution

For first order reaction, half-life is, ( t_ 1 / 2 = 0.693 k ) ( 60= 0.693 k ) ( k= 0.693 60 ) ( t= 2.303 k [A]₀ [A] ) ( 240= 2.303 0.693 60 [A]₀ [A] ) ( [A]₀ [A] = 240 0.693 2.303 60 ) ( [A]₀ [A] =1.20 ) ( a a- = [A]₀ [A] = ) Antilog ( 1.20 ) ( a a-x =15.98 ) ( a=15.98 a-15.98 x ) ( 15.98 x=15.98 a-a ) ( x= 14.98 a 15.98 ) ( =0.9375 a ) ( % ) convertion ( = 0.9375 a 100 a ) ( =93.75 ) The percentage left over ( =100-93.75 )