Question

1) What is the value of Q when each reactant and product is in its standard state? (See Section 10.10 in the textbook for the definition of standard states.)

What is the value of  when each reactant and product is in its standard state? (See Section 10.10 in the textbook for the definition of standard states.)

22.414 L
298 K
1 atm
1

2) For the reactionA(g)⇌2B(g), a reaction vessel initially contains only A at a pressure of PA=1.28 atm . At equilibrium, PA =0.17 atm Calculate the value of Kp. (Assume no changes in volume or temperature.)

3) Consider the reaction: 2NO(g)+Br2(g)⇌2NOBr(g) Kp=28.4 at 298 K
In a reaction mixture at equilibrium, the partial pressure of NO is 102 torr and that of Br2 is 160torr .What is the partial pressure of NOBr in this mixture?

4) The system described by the reaction CO(g)+Cl2(g)⇌COCl2(g)
is at equilibrium at a given temperature when PCO= 0.28 atm , PCl2= 0.10 atm , and PCOCl2= 0.61 atm . An additional pressure of Cl2(g)= 0.40 atm is addedFind the pressure of CO when the system returns to equilibrium.

5)At 650 K, the reaction MgCO3(s)⇌MgO(s)+CO2(g) has Kp=0.026. A 12.5 L container at 650 K has 1.0 g of MgO(s) and CO2 at P = 0.0260 atm. The container is then compressed to a volume of 0.500 L Find the mass of MgCO3 that is formed.

6) Consider the following reaction:
SO2Cl2(g)⇌SO2(g)+Cl2(g)
Kc=2.99×10−7 at 227∘C

If a reaction mixture initially contains 0.168 M SO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

Answer 1

1) what is the value of Q when each reactant and prooduct is in the standard state? The standard state for a substance in a solution is a concentration of 1 M. The standard state for gas is a pressure of 1 atm. therefore Qc= 1 or Qp=1

so answer is D .1

2)

The partial pressure of a gas in a gaseous mixture is proportional to its mole fraction in the mixture.

In this question the partial pressure of A(g) drops from 1.28 atm. to 0.17 atm. as the equilibrium forms.

This reduction (= 1.11 atm.) is caused by molecules of A(g) disappearing as they are converted into B(g).

The equation A(g) <=> 2B(g) tells us that for every molecule of A(g) that disappears, 2 molecules of B(g) are produced.

So, at equilibrium the partial pressure of B(g) must have increased from 0 atm. to 2.22 atm.

So Kp = [pB(g)]^2 / [pA(g)] = 2.22 x 2.22 / 0.17 = 28.99 atm.

3)

2NO + Br2 <----> 2NOBr

Kp = P(NOBr)^2 / P(NO)^2 * (Br2)

102 torr = 0.134 atm

160 torr = 0.210 atm

28.4 = P(NOBr)^2 / (0.134)^2 * (0.210)

28.4 = P(NOBr)^2 /0.00377

P(NOBr) = sqrt(28.4*0.00377) = 0.3272 atm = 248.672 torr

4)

P(CO) = 0.28 atm
P(Cl2) = 0.10 + 0.40 atm = 0.50 atm
P(OCCl2) = 0.61 atm

Since you have more Cl2 than it is required at equilibrium, the reaction will proceed towards the product side diminishing the amounts of Cl2 and CO and increasing the amount of phosgene (OCCl2). Therefore, at equilibrium the concentrations should be:

P(CO) = 0.28 - x
P(Cl2) = 0.50 - x
P(OCCl2) = 0.61 + x

You can write the equilibrium expression for this system:

K = [P(OCCl2)]/{[P(CO)][P(Cl2)]}

Since you know the concentrations at equilibrium (before adding more Cl2) you know what the K value is:

K = (0.61 atm)/{(0.28 atm)(0.10 atm)} = 21.78

Using this value and our expressions for the new pressures at equilibrium gives:

21.78= (0.61 + x)/{(0.28 -x)(0.50 - x)}

21.78{(0.28-x)(0.50-x)} = (0.61 + x)

21.78{(0.14- .28x-0.50x+x^2) = 0.61+x

21.78(x^2 -0.78x+0.14) = 0.61+x

21.78x^2-16.98x+3.04= 0.61+x

21.78x^2 -17.98x +2.43=0

solving equation

x= 0.655 or x= 0.170

x= 0.655 not possible

so x= 0.170

P(CO) = 0.28 - x = 0.28 - 0.17 = 0.11 atm
P(Cl2) = 0.50 - x = 0.50 - 0.17 = 0.33 atm
P(OCCl2) = 0.61 + x = 0.61 + 0.17 = 0.78 atm

so

P(CO) = 0.28 - x = 0.28 - 0.17 = 0.11 atm

5)

the equilibriun has only one gas that contributes to the Kp, that's the CO2:
MgCO3(s) <=> MgO(s) + CO2(g)

Kp = [CO2]

===============

find moles, using molar mass:
1.0 grams Mg) (1 mole MgO / 40.30 grams) = 0.0248 moles of MgO

LeChatelier's principle would entertain the possibliity of shifting the moles of CO2 that had occupied the lost 12.5 Litres & still be able to maintain the 0.026atm & 650 K, which is the only way it could maintain 0.0260 atm of CO2, & its Kp of 0.0260.....
find moles that would need to be shifted to the left, into producing MgCO3......
PV = nRT
(0.0260 atm) (12.5 L) = n (0.08206 L-atm/mol-K) (650K)
n = 0.0060 moles of CO2
that's the limiting reagent

ince MgO(s) + CO2(g react in a 1mole to 1 mole ratio, it is the CO2's moles that determine the amount of MgCO3 that will be produced
by the equation:
MgCO3(s) <= <= <= MgO(s) + CO2(g)
0.0060 moles of CO2 produces an equal 0.0060 moles of MgCO3

using molar mass:
(0.0060 moles of MgCO3) (84.31 g MgCO3 / mole ) = 0.505grams of MgCO3

6)

........... SOCl2 ......... SO2 ....... Cl2
initial ....... 0.168 ....... 0 ....... 0
change........ -x ............+x ........+x
Equil......... 0.168 -x .........x..........x

K = [SO2][O2]/[SO2Cl2]
2.99 x 10^-7 = (x)(x)/(0.168 - x )
neglect x in the 0.168 -x
x^2 = 5.02 x 10^-8

x= 2.24*10^-4

equilibrium concentration of Cl2 = 2.24*10^-4