20. Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbo

Question

20. Methanol (CH3OH) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen:
CO(g)+2H2(g)CH3OH(g).
An equilibrium mixture in a 2.50L vessel is found to contain 2.46×102mol CH3OH, 0.160mol CO, and 0.303mol H2 at 500 K. What is Kc?

[A] 1.7 [B] 1.8 [C] 10.0 [D] 10.5 [E] 10.9

21. The equilibrium
2NO(g)+Cl2(g)2NOCl(g)
is established at 500 K. An equilibrium mixture of the three gases has partial pressures of 9.80×102atm , 0.172atm , and 0.29atm for NO, Cl2, andNOCl, respectively. Calculate Kp for this reaction at 500.0 K.

[A] 45 [B] 48 [C] 49 [D] 51 [E] 55

22. When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibrium, the concentration of N2O4is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction below at 400 K?

N2O4(g) 2 NO2(g)

[A] 0.13 [B] 0.36 [C] 2.5 [D] 0.23 [E] 1.4

23. Dinitrogen tetroxide partially decomposes according to the following equilibrium:

N2O4 (g) 2NO2 (g)

A 1.000-L flask is charged with 3.50 10-2 mol of N2O4. At equilibrium, 2.80 10-2 mol of N2O4 remains. Keq for this reaction is ________

[A] 0.858 [B] 0.427 [C] 2.44 10-4 [D] 0.212 [E] 6.94 10-3

24. Consider the following reaction at equilibrium:

2CO2 (g) INCLUDEPICTURE "http://session.masteringchemistry.com/problemAsset/1801264/3/doublearrowshort.jpg" * MERGEFORMATINET 2CO (g) + O2 (g) H° = -514 kJ

Le Châtelier's principle predicts that removing O2(g) to the reaction container will ________

[A] increase the partial pressure of CO2

[B] decrease the partial pressure of CO

[C] increase the value of the equilibrium constant

[D] decrease the value of the equilibrium constant

[E] decrease the partial pressure of CO2

Explanation / Answer

20)

we know that

concentration = moles / volume (L)

so

[CH3OH] = 2.46 x 10-2 / 2.5 = 9.84 x 10-3

[CO] = 0.16 / 2.5 = 0.064

[H2] = 0.303 / 2.5 = 0.1212

now

Kc = [CH3OH] / [CO] [H2]^2

so

Kc = [9.84 x 10-3 ] / [0.064] [0.1212]^2

Kc = 10.5

so

the answer is D) 10.5

21)

given reaction is

2 NO + Cl2 ---> 2 NOCl

Kp = [pNOCl]^2 / [pNO]^2 [Cl2]

Kp = [0.29]^2 / [9.8 x 10-2]^2 [0.172]

Kp = 50.9

so

the answer is D) 51

22)

we know that

moles = mass / molar mass

so

moles of N204 = 9.2 / 92 = 0.1

now

[N204] = 0.1 / 0.5 = 0.2 M

now

N204 ---> 2 N02

using ICE table

at equilibrium

[N204] = 0.2 - x

[N02] = 2x

given

[N204] = 0.057

so

0.2 - x = 0.057

x = 0.143

so

[N02] = 0.286

now

Kc = [N02]^2 / [N204]

Kc = [0.286]^2 / [0.057]

Kc = 1.435

so

the answer is E) 1.4

23)

[N204]i = 0.035

at

equilibrium

[N204] = 0.035 - x = 0.028

x = 0.007

so

[N02] = 0.014

so

Kc = [0.014]^2 / [ 0.028]

Kc = 7 x 10-3

so

the answer is

E) 6.94 x 10-3

24)

2C02 ---> 2C0 + 02

remove 02

so

the equilibrium will shift in a direction to add 02

so

the equilibrium will shift to the right

so

more and more C02 will be produced

so

partial pressure of C02 is increased

so

option A)

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