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Does Le-Chatelier’s principle predict a change of equilibrium concentrations for the following reaction if the gas mixture is compressed:
N2O4(g)⇌2NO2(g)
One mole of N2O4 gives 2 moles of NO2 gas. The pressure of the gas increases. Upon compressing the gaseous mixture, there is a change in an equilibrium concentration & backward reaction is favoured. This is according to Le Chatlier's Principle
The plot given below shows the relation between (1/T) and log(s), where S is the solubility of an electrolyte AB and T is the temperature in Kelvin. What conclusion can be drawn from the plot:
A mixture of 2 mole each of helium and an unknown gas (normal boiling point = 0°C) is kept in a 22.4 L flask. If the flask is cooled to 0.1°C, the resultant pressure (in atm) inside the flask is:
For the complete oxidation of 100 g of Cyclohexanol to cyclohexanon, the quantity of CrO3 required is (assuming 100% chemical yield) [Atomic wt. of Cr = 52)
As we know that, Cr = 52,
O3 = 16*3 = 48
CrO3 = 52 = 48
= 100 gm
For the reaction: the equilibrium constant at 2000 K and 1.0 bar is 5.25. When the pressure is increased by 8-fold, the equilibrium constant:
In a chemical reaction,
The total pressure at equilibrium is 6 atm. The value of equilibrium constant is:
If the equilibrium constants for the reactions 1 and 2
Are K1 and K2, the equilibrium constant for the reaction
Consider the reaction, at equilibrium,
The equilibrium can be shifted towards the forward direction by:
When a reversible reaction is followed with time with initial concentration of A being 0.6 mol. L–1, the following graph is obtained:
The equilibrium constant K, for the above reaction is:
If the equilibrium reaction is heated, it is observed that the concentration of A increases. Then,
Consider the reaction:
The unit of the thermodynamic equilibrium constant for the reaction is:
An equilibrium constant is defined to be equal to the ratio of the forward and backward reaction rate constants. This particular concentration quotient, , has the dimension of concentration, but the thermodynamic equilibrium constant, K, is always dimensionless.
For the reaction H2(g)+I2(g)⇌2HI (g) at 721 K
value of equilibrium constant is 50, when molar concentration of both hydrogen and iodine is 0.5 M at equilibrium value of Kp under the same conditions will be
The equilibrium reaction is H2(g)+I2(g)⇔2HI(g).
The relationship between Kp and Kc is Kp=Kc(RT)Δn.
For the equilibrium reaction, Δn=2−(1+1)=0.
Hence, Kp=Kc=50.
A 2 L vessel containing 2g of H2 gas at 27°C is connected to a 2L vessel containing 176 g of CO2 gas at 27°C. Assuming ideal behaviour of H2 and CO2, the partial pressure of H2 at equilibrium is………bar.
H2 + CO2 - mixture of non-reacting gases
2g H2 = 1 mol
176g CO2 = 4 mol
Total no. of moles = 5
Total volume = 2+2 = 4L
Temperature = 27+273 = 300K
Total pressure, P=nRT/V=(5*0.083*300/4)= 31.125 bar
According to Dalton's law, p1 = P*(x1)
Mole fraction of H2=1/(1+4)=1/5
Therefore, p(H2)= 31.125*0.2 = 6.225 bar
The relationship between the equilibrium constant K1 for the reaction
And the equilibrium constant K2 for the reaction
The hydrolysis constant (Kh) of NH4Cl is 5.6×10–10. The concentration of H3O+ in a 0.1 M solution of NH4Cl at equilibrium is:
The equilibrium constant Kc for the following reaction at 842°C is 7.90 × 10–3. What is Kp at same temperature:
For the equilibrium what is the temperature at which
At a certain temperature, the following reactions have the equilibrium constants as shown below:
S(s) + O2(g) <=> SO2(g); Kc1 = 5 X 1052
2S(s) + 3O2(g) <=> 2SO3(g) ; Kc2 = 1029
What is the equilibrium constant, Kc for the reaction at the same temperature?
2SO2(g) + O2(g) <=> 2SO3(g)
Equilibrium reactions depend upon the reaction equations
The Kc for the reaction is calculated as,
By the equation,
By reversing the first equation and doubling the coefficient Kc
(KC,)−1
Multiply both side of the equation by K,
2.5×1076
Consider the following gaseous equilibria given below
The equilibrium constant for the reaction, in terms of K1, K2 and K3 will be
Correct Answer :- d
Explanation : K1 = [NH3]2/[N2] [H2]3
K2 = [NO]2/[N2] [O2]3
K3 = [H2O]/[O2]1/2 [H2]
For the reaction :
K = {[NO]2 [H2O]3}/{[NH3]2 [O2]3/2}
= (K2 K33)/K1
9.2 grams of N2O4(g) is taken in a closed on litre vessel and heated till the following equilibrium is reached
At equilibrium, 50% N2O4(g) is dissociated. What is the equilibrium constant (in mol litre–1) (molecular weight of N2O4 = 92)
In a chemical reaction: xenon gas is added at constant volume. The equilibrium:
The equilibrium constants for the reactions and
are K1 and K2, respectively. The equilibrium constant for the reaction
is
Correct statement on the effect of addition of aq. HCl on the equilibrium is:
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