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Chemical Equilibrium
Numericals
SECTION- D: Numericals
Numerical No. 1:
Dinitrogen tetra oxide N2O4, decomposed into nitrogen dioxide NO2, in a reversible reaction. Derive equilibrium constant expression for the reaction of decomposition. Also interpret unit of Kc for balanced chemical reversible reaction.
Solution:
Equation:
N2O4 ⇌ 2NO2
Derivation Of Equation For Equilibrium Constant:
Let us apply law of mass action on above reversible reaction.
Forward Reaction:
The rate of forward reaction according to law of mass action is:
Rf ∝ [N2O4]
Rf = kf [N2O4] ...(i)
Where k
f is the rate constant for forward reaction.
Reverse reaction:
The rate of reverse reaction according to law of mass action is:
Rr ∝ [NO2]2
Rr = kr [NO2]2 ... (ii)
Where k
r is the rate constant for reverse reaction.
At Equilibrium:
We know that, at equilibrium rate of forward and reverse reaction becomes equal. So,
Rf = Rr ... (iii)
Putting the values of R
f and R
r from eq. (i) & (ii) in eq. (iii), we have:
kf [N2O4] = kr [NO2]2
By taking constants on L.H.S and variables on R.H.S, we have
Where K
c is called equilibrium constant.
Unit Of Kc for given reaction:
Ans: Hence Unit of K
c for the given reaction is [mol.dm
-3].
Numerical No. 2:PCl5, PCl3 and Cl2 are at equilibrium at 500K in a closed container and their concentrations are 0.8 x 10-3 mol.dm-3, 12 x 10-3 mol.dm-3 and 1.2 x 10-3 mol.dm-3 respectively. Calculate the value of Kc for the reaction along with unit.
Solution:
Given:-
[PCl5] = 0.8 x 10-3 mol.dm-3
-
[PCl3] = 12 x 10-3 mol.dm-3
-
[Cl2] = 1.2 x 10-3 mol.dm-3
-
Kc at 500K = ?
Equation:PCl5(g) ⇌ PCl3(g) + Cl2(g)
Therefore
Ans: The value of K
c at 500K is 1.8 x 10
-3 mol.dm
-3
Numerical No.3:
The value of Kc for the reaction is 1 x 10-4
2HI(g) ⇌ H2(g) + I2(g)
At a given temperature, the molar concentration of reaction mixture is HI = 2 X 10-5 mol.dm-3, H2 = 1 X 10-5 mol.dm-3 and I2 = 1 x 10-5 mol.dm-3.
Predict the direction of the reaction.
Solution:
Given:
-
Kc = 1 x 10-4
-
[HI] = 2 X 10-5 mol.dm-3
-
[H2] = 1 X 10-5 mol.dm-3
-
[I2] = 1 x 10-5 mol.dm-3
-
Direction of reaction = Qc = ?
Working:
At a given temperature the reaction quotient Qc for the reaction will be given by the expression: As the value of the reaction quotient is greater than the value of Kc i.e 0.25 > 1 x 10-4
Ans: Hence Qc > Kc, the reaction will be proceed in reverse direction.
Test Yourself
Numerical No. 4:
The value of Kc for the following reaction at 717K is 48.
H2(g) + I2(g) ⇌ 2HI(g)
At a particular instant, the concentration of H2, I2 and HI are found to be 0.2 molL-1, 0.2 molL-1, and 0.6 molL-1 respectively. Calculate reaction quotient for given reaction. Also predict direction of reaction.
Solution:
Given:
-
Kc = 48 at 717K
-
[H2] = 0.2 molL-1
-
[I2] = 0.2 molL-10.2 molL-1
- [HI] = 0.6 molL-1
Working Formula:
At a given time the reaction quotient Qc for given reaction will be given by the expression:
Therefore Qc < Kc
Ans: Qc < Kc, so reaction will proceed in forward direction.
Numericals - Examples From Text Book
Numerical 01:
Equilibrium occurs when nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide gas.
2NO(g) + O2(g) ⇄ 2NO2(g)
At equilibrium at 230°C, the concentrations are measured to be [NO] = 0.0542 mol.dm-3, [O2] = 0.127 mol.dm-3, and [NO2] = 15.5 mol.dm-3. Calculate the equilibrium constant at this temperature.
Solution:
Given equilibrium concentrations of reactants and product are:
-
[NO] = 0.0542 mol.dm-3
-
[O2] = 0.127 mol.dm-3
-
[NO2] = 15.5 mol.dm-3
Write equilibrium expression as:
Now put equilibrium concentration values in equilibrium constant expression
Ans: Kc = 6.44 X 105 mol-1.dm3
Numerical 02:
A reaction takes place between iron ion and chloride ion as:
Fe+3(aq) + 4Cl-(aq) ⇌ FeCl4-(aq)
At equilibrium, the concentrations are measured to be (Fe+3) is 0.2 mol.dm-3, Cl- is 0.28 mol.dm-3 and FeCl4- is 0.95x10-4 mol.dm-3. Calculate equilibium constant Kc for given reaction.
Solution:
Given equilibrium concentrations of reactants and product are:
-
[Fe+3] = 0.2 mol.dm-3
-
[C1- = 0.28 mol.dm-3
-
[FeCl4-] = 0.95x 10-4 mol.dm-3
Write equilibrium expression as:
Now put equilibrium concentration values in equilibrium constant expression:
Ans: Kc = 7.72 x 10-2 mol-4.dm12
Numerical 03:
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperature. At 2000 °C, the value of the equilibrium constant for the given reaction is 4.1 x 10-4
N2(g) + O2(g) ⇌ 2NO(g)
Find the concentration of NO in an equilibrium mixture at 1 atm pressure at 2000°C. In air, [N2 = 0.036 mol/L and [O2] = 0.0089 mol/L
Solution:
Given: equilibrium concentrations of reactants and product are:
-
[N2] = 0.036 mol/L
-
[O2] = 0.0089 mol/L
-
[NO] = ?
We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing
Taking square root on both the sides, we have
√[NO2] = √(4.1 x 10-4 mol/L)(0.036 mol/L) (0.0089 mol/L)
Ans: [NO] = 3.6x10-1 mol/L