I.
1-lenta
2-gazoasa
3-nu se consuma
4-scade
5-biocatalizatori
II.
1-a
cinetica de ordinul I
=> v = k[A] = 5x10la-2x1,5 = 7,5x10la-2 mol/L*s
2-d
3-c
4-b
5-c
t = 10 min = 600 s
stim ca v.mediu = -Δc/Δt = (1,06-2)/(600-0)
= 1,57x10la-3 mol/L*s
III.
nr. moli.H2 = 0,08/2 = 0,04 moli
nr.moliI2 = 1/127 = 7,87x10la-3 moli
H2 + I2 <==> 2HI
v = kx[H2]x[I2]
[H2] = 0,04/1 = 0,04 mol/L
[I2] = 7,87x10la-3/1 = 7,87x10la-3 mol/L
=> v = 0,42x0,04x7,87x10la-3 = 1,32x10la-4 mol/Lxs
IV.
pentru ecuatia a)
v = k[C]²[H2]² => [H2]a = √(va/k[C]²a)
pentru ecuatia b)
v = k[C]²[H2] => [H2]b = vb/k[C]b
√(va/k[C]²a) = vb/k[C]b , va=vb
=> √(1/[C]²a) = 1/[C]b
=> [H2]a/[H2]b = [C]a/[C]b
