Question

The first-order rate constant for the decomposition of N2O5, N2O5(g) 2NO2(g) + O2(g)At 70C is 6.810-3s-1. Suppose we start with 0.0250 mol of N2O5(g) in a volume of 1.0 L. a.) How many moles of N2O5will remain after 2.5 min? b.)How many minutes will it take for the quantity of N2O5to drop to 0.010 mol? c.What is the half-life of N2O5at 70 degrees C?

Answer 1

a) For the first order reaction

$ln(\frac{A_0}{A_t}) = kt$

$ln(\frac{A_0}{A_t}) = 6.8 * 10^{-3} * 2.5 * 60 = 1.02$

$A_t =\frac{A_0}{e^{1.02}} = 0.00901 \ moles$

b)

$ln(\frac{A_0}{A_t}) = kt$

$ln(\frac{0.0250}{0.010}) = 6.8 * 10^{-3} * t$

$t = \frac{1000 * ln(2.5)}{6.8} = 134.74s$

c)

$t_{1/2} = \frac{ln2}{k} = \frac{1000 * ln(2)}{6.8} = 101.9s$