Equilibrium Constant and Standard Free Energy Change
Trending Questions
- HI⇌12H2+12I2
- PCl5⇌PCl3+Cl2
- N2O4⇌2NO2
- all the above
At 298 K and 1 atm pressure, the partial pressures in an equilibrium mixture of N2O4 and NO2 are 0.7 and 0.3 atmosphere respectively. What will be the partial pressure of NO2 when the gases are in equilibrium at 298 K and at a total (equilibrium) pressure of 10 atmospheres?
2.23 atm
0.77 atm
1.078 atm
1.60 atm
- 0oC
- 273 K
- 1 K
- 12.18 K
What do you understand by products
- Low pressure
- High pressure
- High temperature
- Catalyst
- The pressure inside the container increases
- The temperature will decrease
- The pressure inside the container decreases
- The temperature will remain same
PCl5(g)⇌PCl3(g)+Cl2(g)
Also, calculate the standard entropy change if,
ΔH∘=28.40 kJ mol−1
- 38.48 kJ, −33.6 J mol−1K−1
- 48.48 kJ, −33.6 J mol−1K−1
- 38.48 kJ, −43.6 J mol−1K−1
- 48.48 kJ, −43.6 J mol−1K−1
Which of the following statements is correct for a reversible process in a state of equilibrium ?
ΔG∘=2.30 RT log K
ΔG=−2.30 RT log K
ΔG=2.30 RT log K
ΔG∘=−2.30 RT log K
2N2O(g)⇌2N2(g)+O2(g)
ΔG0f(N2O)=104.2 kJ/mol, ΔG0f(N2)=0 and ΔG0f(O2)=0
- 6.3×1030
- 3.3×1036
- 3.3×10−36
- 6.3×10−30
In the conversion of lime stone to lime, CaCO3(s) → CaO(s) + CaO2(g) the values of ΔH∘ and ΔS∘ are +179.1 kJ mol−1 ad 160.2 JK−1 mol−1 respectively at 298 K and 1 bar. Assuming, ΔH∘ and ΔS∘ do not change with temperature; temperature above which conversion of lime stone to lime will be spontaneous is
1118 K
1008 K
1200 K
845 K
Which is correct statement if N2 is added at equilibrium condition?
- The equilibrium will shift to forward direction because according to IInd law of thermodynamics, the entropy must increases in the direction of spontaneous reaction
- The condition for equilibrium is G(N2)+3G(H2)=2G(NH3) where, G is Gibbs free energy per mole fo the gaseous species measured at that partial pressure. The condition of equilibrium is unaffected by the use of catalyst, which increases the rate of both the forward and backward reactions to the same extent
- The catalyst will increase the rate fo forward reaction by α and that of backward reaction by β
- Catalyst will not alter the rate fo either fo the reaction
- ΔU=0, q=2370J, w=−23.70×102J
- ΔU=2000, q=23715, w=24.70×102J
- ΔU=0, q=2370J, w=−24.70×102J
- ΔU=0, q=2374J, w=−23.70×102J
- For exothermic reaction, van't Hoff plot would always gives positive slope
- For exothermic reaction, van't Hoff plot would always gives negative slope
- For exothermic reaction, van't Hoff plot can gives both positive and negative slope
- None of the above
- 20.16
- 2.303
- 2.016
- 13.83
N2O4 (g)⇌2NO2 (g)
(GoN2O4)298K=100 kJ mol−1(GoNO2)298K=50 kJ mol−1
If 5 moles of each species is taken in a 1 litre container. Then, calculate the value of ΔG at 298 K and predict the spontaneity of the reaction .
- −2.5 kJ and spontaneous.
- +2.5 kJ and non-spontaneous.
- +4 kJ and non-spontaneous.
- −4 kJ and spontaneous.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a)
(b)
(c)