Equilibrium
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New answer posted
5 months agoContributor-Level 10
Let x moles of BrCl decompose in order to attain the equilibrium. The initial molar concentration and the molar concentration at equilibrium point of different species may be represented as follows:
2BrCl2?(g) ?→?Ar2?(g)+Cl2?(g)
3.3*10−3m 0 0
3.3*10−3−2αα? α
∴ Kc?= (α*α)? /(3.3*10−3−2α)2
⇒ α2 / (3.3*10−3−2α)2?=32
⇒ α / (3.3*10−3−2α)2?=4
⇒ α=18.67*10−3−8 α
⇒ (1+ 8 α =18
New answer posted
5 months agoContributor-Level 10
According to available data:
N2 = [3.0], H2 = [2.0], NH3 = [0.50]
Qc = [NH3]2 / [N2] [ H2]3
= [0.50]2 / [3.0] [ 2.0]3
= 0.25/24
= 0.0104.
Since the value of Qc is less than that of Kc (0.061), the reaction is not in a state of equilibrium. It will proceed in the forward direction till Qc becomes the same as Kc.
New answer posted
5 months agoContributor-Level 10
The initial partial pressures of CO and CO2 are 1.40 atm and 0.8. atm respectively.
The expression for the reaction quotient is:
Qp=PCO2 / PCO=0.80 / 1.4 = 0.571
As the value of the reaction quotient is greater than the value of the equilibrium constant, the reaction will move in the backward direction.
To attain the equilibrium, the partial pressure of CO2 will decrease and the partial pressure of CO will increase.
Let p atm be the decrease in the partial pressure of CO2. The in
New answer posted
5 months agoContributor-Level 10
Let the initial molar concentration of PCl5 per litre = x mol
Molar concentration of PCl5 at equilibrium = 0.05 mol
Therefore, moles of PCl5 decomposed = (x – 0.05) mol
Moles of PCl3 formed = (x – 0.05) mol
Moles of Cl2 formed = (x – 0.05) mol
The molar conc./litre of reactants and products before the reaction and at the equilibrium point are:
| PCl5 | PCl3 | Cl2 |
Initial moles/litre | x | 0 | 0 |
At equilibrium | 0.05 | x – 0.05 | x – 0.05 |
Kc = [PCl3] [Cl2] / [PCl5]
=> 8.3 x 10-3 = (x – 0.05)2 / 0.05
=> (x – 0.05) = (4.15 x 10-4)1/2 = 2.037 x 10-2 = 0.02 moles
=> x = 0.05 + 0.02 = 0.07 mol
Therefore, at equilibrium:
Moles of PCl3 formed = (x – 0.05) mol = 0.07 – 0.05 = 0.02 mol
Moles of C
New answer posted
5 months agoContributor-Level 10
(i) The concentration ratio (Concentration quotient), Qc for the reaction is:
Qc = [CH3COOC2H5] [ H2O] / [CH3COOH] [C2H5OH]
(ii)
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O |
Initial molar concentration | 1.0 mol | 0.18 mol | 0 | 0 |
Molar concentration at equilibrium | (1 – 0.171) = 0.829 mol | (0.18 – 0.171) = 0.009 mol | 0.171 mol | 0.171 mol |
Applying
Kc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
= (0.171 mol) x (0.171 mol) / (0.829 mol) (0.009 mol)
= 3.92
(iii)
| CH3COOH | C2H5OH | CH3COOC2H5 | H2O |
Initial molar concentration | 1.0 mol | 0.5 mol | 0.214 mol | 0.214 mol |
Molar concentration at equilibrium | (1 – 0.214) = 0.786 mol | (0.5 – 0.214) = 0.286 mol | 0.214 mol | 0.214 mol
|
Qc = [CH3COOC2H5] [H2O] / [CH3COOH] [C2H5OH]
= (0.214 mol) x (0.214 mol) / (0.786 mol) (0.286 mol)
= 0.204
Since Qc is less than Kc, this means that the equilibrium has not been reached. The reactants are still taking part in the reaction to form the products.
New answer posted
5 months agoContributor-Level 10
The equilibrium reaction is
C2?H6?(g) ? C2?H4?(g)+H2?(g).
Initial | 4 | 0 | 0 |
Change | −x | x | x |
Equilibrium | 4−x | x | x |
The expression for the equilibrium constant is Kp?= (?PC2?H4?) (PH2) / PC2?H6?.
Substituting the values in the above equation, we get
0.04=x2 / (4−x)
? or x=0.38
Thus, the pressure of ethane is, PC2?H6?=3.62atm.
New answer posted
5 months agoContributor-Level 10
Suppose at equilibrium, the molar concentration of both I2 and Cl2 is x mol L-1.
Kc = [I2] [ Cl2] / [ICl]2= x2 / (0.78 – 2x)2
=>x/ (0.78 – 2x) = (0.14)1/2 = 0.374
=> x= 0.167
[ICl] = (0.78 – 2 x 0.167) = (0.78 – 0.334) = 0.446 M
[I2] = 0.167 M,
[Cl2] = 0.167 M
New answer posted
5 months agoContributor-Level 10
H2 (g) + I2 (g)? 2HI (g); K=64, T=700K
2HI? H2 + I2 K=1/64700K
a 0 0
a (1−α) aα/2 aα/2
0.5 x x
x2 / (0.5)2 = 1 / 54.8
x2 = 0.25 / 54.8
x = = 0.068 M
At equilibrium, [H2] = [I2] = 0.068 M
New answer posted
5 months agoContributor-Level 10
Number of moles of water originally present = 1 mol
Percentage of water reacted= 40%
Number of moles of water reacted= 1 x 40/100 = 0.4 mol
Number of moles of water left= (1 – 0.4) = 0.6 mole
According to the equation, 0.4 mole of water will react with 0.4 mole of carbon monoxide to form 0.4 mole of hydrogen and 0.4 mole of carbon dioxide.
Thus, the molar conc., per litre of the reactants and products before the reaction and at the equilibrium point are as follows:
| H2O | CO | H2 | CO2 |
Initial moles / litre | 1/10 | 1/10 | 0 | 0 |
At Equilibrium | (1 – 0.4) / 10 = 0.6/10 | (1 – 0.4) / 10 = 0.6/10 | 0.4/10 | 0.4/10
|
Equilibrium constant, Kc= [H2] [CO2] / [H2O] [CO]
= [ (0.4/10) x (0.4/10)] / [ (0.6/10) x (0.6/10)]
= 0.16 / 0.36 = 0.44
New answer posted
5 months agoContributor-Level 10
Balanced chemical equation for the reaction is
4 NO (g) + 6 H2O (g)? 4 NH3 (g) + 5 O2 (g)
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