For a first-order gas-phase decomposition reaction, the half-life is 15 minutes. If the initial partial pressure of the reactant is 800 ~mm~Hg , what will be its partial pressure after 60 minutes?

Step-by-step solution

For a first-order reaction, the amount of reactant remaining after n half-lives is given by P_t = P₀ 2^n . First, determine the number of half-lives ( n ) that have elapsed: n = Total time Half-life = 60 minutes 15 minutes = 4 Now, calculate the partial pressure after 4 half-lives: P_t = 800 ~mm~Hg 2^4 P_t = 800 16 = 50 ~mm~Hg Dividing the initial pressure by n ( 800/4 = 200 ) or by 2n ( 800/8 = 100 ) are common mistakes resulting from not using the exponential relationship. Answer: 50 ~mm~Hg