first order reaction has rate constant 0.0051min-1 if we begen with 0.01 molar conc.of reactants what conc. of reactants will remain in solution after 3 hour

Question

Antilog 0.3986=2.503

Answer 1

For first order reaction, t_1/2 = ln2/k = 0.693/k = 0.693/0.0051 min^-1 = 135.9 min. 

Also, for first order reaction, fraction remaining at time t = 0.5ⁿ where n = number of half lives elapsed.

Number of half lives elapsed = 3 hours x 60 min/hour x 1 half live/135.9 min = 1.3245 half lives elapsed

Fraction remaining = 0.5^1.3245 = 0.399

Concentration of reactant remaining = 0.399 x 0.01 M = 0.00399 M

 

You should get the same answer if you use the form of the integrated rate equation [A]/[A]_o = e^-kt

Substitute (0.0051 min^-1 x 180 min) = 0.918 as the value of kt and solve for [A]