Question

The deposition of HOF occuras at 25C 2 HOF(g) = 2 HF(g) +02 (g) I need to show my work and all equations used. I also included the answers but I do not know how to calculate them

The graph is attached at   

https://docs.google.com/document/d/10__3VzL8YDJaqchUPYVmSfhQyH5fYMts9Oj9jMRjs4g/edit?usp=sharing

What is the initial concentration of HOF? .54

Write the rate law for this reaction. Include the value and units for the rate constant. rate .17 1[HOF}

How long will it take for the concentration of HOF to decrease to 27% of its initial concentration in seconds? 4.6 x 10^1s

What is the half-life for the decomposition of HOF at 25C in seconds? 2.4 x 10^1

Thank you

Answer 1

Initial concentration of HOF = inv.ln.(-0.15) = 0.84 M [first value from the top on y-axis)

pl. note If initial value changes, all the rest of calculation values also change accordingly.

Since plot of ln[HOF] vs time gives us a straight line, this is a first order reaction.

Rate law would be, rate = k[HOF], first order reaction kinetics

For a first order reaction, ln[HOF] = ln[HOF]o - kt

from graph, slope = -k = -0.017

k = 0.017 s-1

ln(27) = ln(100) - 0.017 x t

Thus time taken for initial concentration to reduce to 27% (t) = 77.02 s

hald life, t1/2 = 0.693/0.017 = 40.76 s

answered by: anonymous