The rate law of the reaction NH3 + HOCl → NH2Cl + H2O is rate = k[NH3][HOCl] with k = 5.1 × 106 L/mol·s at 25°C. The reaction is made pseudo-first-order in NH3 by using a large excess of HOCl. How long will it take for 40% of the NH3 to react if the initial concentration of HOCl is 2 × 10−3 M?

40% of the ammonia will take 4.97x10^-5 s to react.

Explanation:

The rate is equal to:

R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]

R = k´ * [NH3]

k´ = 10200 s^-1

Because k´ is the psuedo first-order rate constant, we have the following:

b/(b-x) = 100/(100-40) ; 40% ammonia reacts

b/(b-x) = 1.67

log(b/(b-x)) = log(1.67)

log(b/(b-x)) = 0.22

the time will equal to:

t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s

h let me think

adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: