Question

1.Chemical kinetics

4NH3 + 3O2 = 2N2 + 6H2O

Given, rate of N2 = 0.45 Moles /Ls Calculate the rate of NH3?

2. The thermal composition of phosphine (PH3) into phosphorus and molecular hydrogen is a first order reaction:

4PH3(g) P4(g) + 6H2(g)

The half life of a reaction is 35 sec at 6800C. Calculate the time required for 80% of the phosphine to decompose.

Answer 1

1) 4 NH3 + 3 02 --> 2 N2 + 6 H20

we can see that

(1/4) (-d[NH3] / dt) = ( 1/2) (d[N2] / dt)

so

(1/4) (-d[NH3] / dt) = (1/2) (0.45)

-d[NH3]/dt = 0.45 x 4 / 2 = 0.9

d[NH3]/dt = -0.9

so

the rate of NH3 is -0.9 moles / Ls

2)

now for 1st order reaction

A = Ao x e^(-kt)

A/Ao = e ^(-kt)

ln (A/Ao) = -kt

given 80% decomposed

so 20% is left

so A/Ao = 0.2

now

ln 0.2 = -kt

kt = 1.609438

now

t = 1.609438 / k

we know that

k = ln 2 / t1/2

so

t = (1.609438 / ln 2) x t1/2

t = (1.609438 / ln2 ) x 35

t = 81.267 sec

so

the time required is 81.267 sec