Afirst-order reaction has a half-life of 29.2 s . how long does it take for the concentration of the reactant in the reaction to fall to one-sixteenth of its initial value?

Answer is: it takes 116,8 seconds to fall to one-sixteenth of its initial value

The half-life for the chemical reaction is 29,2 s and is independent of initial concentration.
c₀ - initial concentration the reactant.

c - concentration of the reactant remaining at time.

t = 29,2 s.
First calculate the rate constant k:
k = 0,693 ÷ t = 0,693 ÷ 29,2 s = 0,0237 1/s.
ln(c/c₀) = -k·t₁.
ln(1/16 ÷ 1) = -0,0237 1/s · t₁.

t₁ = 116,8 s.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :