1. Determine the equilibrium expression for the following reaction 2 NaHCO 3 (s)
ID: 976660 • Letter: 1
Question
1. Determine the equilibrium expression for the following reaction
2 NaHCO3 (s) Na2CO3 (s) + H2O (g) + CO2 (g)
2. Kc = 5.6 × 10-12 at 500 K for the reaction
I2 (g) 2 I (g)
A mixture is kept at 500 K and before equilibrium is reached, I2 has a concentration of 0.02 M and I has a concentration of 0.00000002 M. Which direction must the reaction lie to reach equilibrium.
3. Given Kp = 1.5 × 1018 at 300 K for the reaction below
3 NO N2O + NO2
a. Determine the Kc value at the same temperature.
b. 900 mg of NO were initially placed in a 1.00 L vessel and the reaction was allowed to reach equilibrium. Calculate the equilibrium concentrations of NO, N2O, and NO2.
4. Calculate the Kc for the following reaction
SnO2 + 2CO Sn + 2 CO2
given the following information:
SnO2 + 2 H2 Sn + 2 H2O Kc = 8.12
H2 + CO2 CO + H2O Kc = 0.771
5. Methanol is manufactured by the below catalyzed reaction using ZnO/Chromate.
CO (g) + 2 H2 (g) CH3OH (g) D H = -91 kJ
Does the amount of methanol (CH3OH) increase, decrease, or remain the same when an equilibrium mixture of reactants and/or products are subjected to the following changes?
a. The temperature is decreased
b. The volume is increased
c. Methanol is added
d.
Explanation / Answer
1. Determine the equilibrium expression for the following reaction
2 NaHCO3 (s) --- > Na2CO3 (s) + H2O (g) + CO2 (g)
Keq = PCO2 PH2O
Here NaHCO3 (s) and Na2CO3 (s) both present in solid form.
2. Kc = 5.6 × 10-12 at 500 K for the reaction
I2 (g) --- > 2 I (g)
A mixture is kept at 500 K and before equilibrium is reached, I2 has a concentration of 0.02 M and I has a concentration of 0.00000002 M. Which direction must the reaction lie to reach equilibrium.
Now determined the Q fro the given reaction:
Q = product / reactants
Q= 0.00000002 / 0.02
Q=1.0*10^-6
And given that Kc = 5.6 × 10-12 at 500 K for the reaction
Here Q, 1.0*10^-6 > Kc, 5.6 × 10-12
When Q>K, there are more products than reactants. The direction of reaction is forward .
3. Given Kp = 1.5 × 1018 at 300 K for the reaction below
3 NO --- > N2O + NO2
The relation between Kp and Kc is given as;
Kp = Kc (RT)^n
Kp = 1.5 × 1018 at 300 K
Kp = equilibrium constant in terms of partial pressures (atm)
Kc = equilibrium constant in terms of concentrations(molarity)
R = ideal gas constant ( 0.082 atm.L / K.mol)
T = absolute temperature (300 K)
n = n products - n reactants
n = (2) - (3) =- 1
Kp = Kc (RT)^n
1.5 × 1018 at 300 K = (Kc) {(0.082)(300)}^-1
Kc = 3.69*10^19
b. 900 mg of NO were initially placed in a 1.00 L vessel and the reaction was allowed to reach equilibrium. Calculate the equilibrium concentrations of NO, N2O, and NO2.
4. Calculate the Kc for the following reaction
SnO2 + 2CO ---> Sn + 2 CO2
given the following information:
Given that,
SnO2 (s) + 2H2 (g) <--> Sn (s) + 2H2O (g)
Kc1= 8.12 = [H2O]^2 / [H2]^2
H2 (g) + CO2 (g) <--> H2O (g) + CO (g)
Kc2 = 0.771 = [H2O] [CO] / [H2] [CO2]
To calculate the Kc for the following reaction
SnO2 + 2CO ---> Sn + 2 CO2
Kc = [CO2]^2 / [CO]^2
first we take the inverse squared of Kc2,
(1/ Kc2)^2 = [H2]^2 [CO2]^2 / [H2O]^2 [CO]^2
Now the above equation is multiply with Kc1
(1/ K2)^2 X Kc1 =( [H2]^2 [CO2]^2 / [H2O]^2 [CO]^2 ) X ([H2O]^2 / [H2]^2 )
= CO2]^2 / [CO]^2
The above is the result of the given reaction.
K = (1/ K2)^2 X Kc1 = (1/ 0.771)^2 x (8.12)
K = 13.66
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