1) For a certain reaction, K c = 9.59×10 4 and k f= 32.6 M 2s1 . Calculate the v
ID: 882629 • Letter: 1
Question
1) For a certain reaction, Kc = 9.59×104 and kf= 32.6 M2s1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction.
For a different reaction, Kc = 0.564, kf=52.3s1, and kr= 92.7 s1 . Adding a catalyst increases the forward rate constant to 2.62×104 s1 . What is the new value of the reverse reaction constant, kr, after adding catalyst?
2)Consider the reaction:
N2(g)+3H2(g)2NH3(g)
Complete the table. Assume that all concentrations are equilibrium concentrations in moles per liter, M.
Complete the first row of the table.,Complete the second row of the table.Complete the third row of the table.
3)Consider the reaction:
SO2Cl2(g)SO2(g)+Cl2(g)
A solution is made containing initial [SO2Cl2]= 0.024 M . At equilibrium, [Cl2]= 1.2×102 M .
Calculate the value of the equilibrium constant.
Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of SO2Cl2 and SO2.
4)An equilibrium mixture of the following reaction was found to have [SO3]= 0.341 M and [O2]= 0.165 M at 600 C.
What is the concentration of SO2?
2SO2(g)+O2(g)2SO3(g)
Keq= 4.84 at 600 C
4)
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M , [B] = 0.800 M , and [C] = 0.550 M . The following reaction occurs and equilibrium is established:
A+2BC
At equilibrium, [A] = 0.510 M and [C] = 0.690 M . Calculate the value of the equilibrium constant, Kc.
Express your answer numerically.
T(K) [N2] [H2] [NH3] Keq 500 0.120 0.105 0.459 _____ 575 0.120 _____ 0.128 8.6 775 0.100 0.175 _____ 0.0434Explanation / Answer
Solution
1) We know that
Kc = kf/kr
kr = kf / Kc = (32.6 M-2s-1)/(9.59×104)
kr = 3.39 x 104M-2s-1 (ANS)
After adding catalyst, the rate of forward reaction increases nut the equilibrium constant does not change. Therefore,
Kc = 0.564,
Kf = 2.62×104 s1 (after catalyst addition)
Thus, kr = kf / Kc = 2.62×104 s-1/0.564
kr =4.64 ×104 s-1 (ANS)
2)
T(K)
[N2]
[H2]
[NH3]
Keq
500
0.120
0.105
0.459
1516.78
575
0.120
0.251
0.128
8.6
775
0.100
0.175
_0.00275
0.0434
(i) The first row of the table
Kc = [NH3]2/[N2] [H2]3
= (0.459)2/(0.120)(0.105)3
= 0.210681/0.0001389
= 1516.78
(ii) The second row of the table
8.6 = [0.128]2/[0.120] [H2]3
[H2]3= 0.016384/1.032 = 0.0159
[H2] = 0.2514
(iii) The third row of the table
0.0434 = [NH3]2/ (0.175)(0.100)3
[NH3]2 = 0.175 x (0.100)3x 0.0434 = 0.000007595
[NH3] = 0.00275
3) For the reaction
SO2Cl2(g) SO2(g) + Cl2(g)
[SO2Cl2]
[SO2]
[Cl2]
I
0.024
0
0
C
-1.2×102
+1.2×102
+1.2×102
E
0.024-1.2×102
1.2×102
1.2×102
Kc = [SO2][Cl2] / [SO2Cl2]
Kc = (1.2×102 x 1.2×102)/( 0.024-1.2×102)
= 0.000144/0.012
= 0.012 (ANS)
T(K)
[N2]
[H2]
[NH3]
Keq
500
0.120
0.105
0.459
1516.78
575
0.120
0.251
0.128
8.6
775
0.100
0.175
_0.00275
0.0434
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