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?

Question

Chemistry

Maliyah Peterson

1 November, 18:50

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?

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Atticus Anthony · Answer 1

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