1 2 100 50 0 50 0.20 0.20 0.40 0.40 67 67 33 33 1 2 200 50 0 150 0.60 0.60 0.20
ID: 539253 • Letter: 1
Question
1
2
100
50
0
50
0.20
0.20
0.40
0.40
67
67
33
33
1
2
200
50
0
150
0.60
0.60
0.20
0.20
50
50
150
150
1
2
80
59
20
41
0.20
0.20
0.60
0.60
75
75
24
24
a) what is the distinction between [A]0 and [A] in the colum headings?
d) examine the table above. if kr<kr what are the relative values of the equilibrium concentrations of A and B? Explain. if kf >kr what are the relative values of the equilibrium concentrations of A and B? Explain.
f) show that [B]/[A] is constant
Set Trial [A]0 [B]0 kf kr [A] [B] M1
2
100
50
0
50
0.20
0.20
0.40
0.40
67
67
33
33
N1
2
200
50
0
150
0.60
0.60
0.20
0.20
50
50
150
150
O 1 100 0 0.50 0.50 50 50 P1
2
80
59
20
41
0.20
0.20
0.60
0.60
75
75
24
24
Explanation / Answer
A(g) <=======>B(g)
Where rate forward=kf[A]
Rate reverse =kr[B]
a) In the column headings [A]0 is initial concentration of reactant A while [A] is remaining concentration of reactant A ,after the product B is formed.
b) For trial M2,
The rate of the forward reaction = kf[A]
= 0.20 x 67 = 13.40
The rate of reverse reaction = kr[B]
= 0.60 x 33 =13.20
At equilibrium and the rate of the forward reaction is approximately equal to rate of reverse reaction
c) For trial P2
The rate of the forward reaction = kf[A]
= 0.20 x 75 = 15.00
The rate of reverse reaction = kr[B]
= 0.60 x 24 =14.4 0
At equilibrium and the rate of the forward reaction is slightly higher than rate of reverse reaction.
d) If kr<kr the relative values of the equilibrium concentrations of [A] is low and [B ] is high.
Explanation- If rate of reverse reaction is low, then equilibrium sift to right .More product is formed and concentration of [B] is high.
if kf >kr the relative values of the equilibrium concentrations of [A] is low and [B ] is high
Explanation - If rate of forward reaction is more than reverse reaction, then equilibrium sift to right and more product is formed and concentration of [B] is high. The increase in the forward rate means that the system is not in equilibrium and the concentration of reactant [A] decreases, while that of product [B] increases.
e) In general, the rate of the forward reaction = reverse reaction at equilibrium. When the two rates are equal there is no further change in the concentration of the reactants or products, although both reactions continue, and the system has attained dynamic equilibrium.
f)
At equilibrium the reaction has reached a point where the concentrations of the reactant and product are not changed with time, because the forward and backward reactions have the same rate.
Keq = [B]/[A] )
So, [B]/[A] is constant
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