The first order gas reaction $A (g) 2 B (g)$ starts with pure $A$ at 80 mmHg. After 25 min the total pressure is 110 mmHg. Find $k$ in units of $1 0^{- 2} mi n^{- 1}$ (two decimals; use $ln 1.6 = 0.470$ ).

Numerical answer type, answer in 10^-2 min^-1, tolerance 0.02.

Answer: 1.88 10^-2 min^-1

x

of A reacts,

p_{A} = 80 - x

and

P_{t} = 80 + x

, so

p_{A} = 2 P_{0} - P_{t} = 160 - 110 = 50

mmHg. Then

k = \frac{1}{t} ln \frac{P _{0}}{p _{A}} = \frac{ln ( 80/50 )}{25} = \frac{0.470}{25} = 0.0188 mi n^{- 1}

, i.e.

1.88 \times 1 0^{- 2}