Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant
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ΔG∘ for the reaction x + y ⇌ z is −4.606 kcal. The value of equilibrium constant of the reaction at 227∘C is
100
10
2
0.01
For the reaction 2NO(g)⇌N2(g)+O2(g), ΔH=−180 kJ mol−1. Which of the following facts does not hold well?
The dissociation of NO is favored more at high temperature
The pressure changes at constant temperature do not affect the equilibrium constant
The volume changes at constant temperature do not affect the equilibrium constant
The dissociation of NO is favored less at high temperature
For a reaction ΔH = 9.08 kJ mol−1 and ΔS = 35.7 JK−1 mol−1 Which of the following statements is correct for the reaction
Reversible and Isothermal
Reversible and Exothermic
Spontaneous and Endothermic
Spontaneous and Exothermic
- H2(g)+I2(g)⇌2HI(g) ΔrH=+3000 cal
- N2(g)+O2(g)⇌2NO(g) ΔrH=−43200 cal
- 2NH3(g)⇌N2(g)+3H2(g) ΔrH=+22400 cal
- CO2(g)⇌C(s)+O2(g) ΔrH=+94300 cal
- Nature of reactant and product.
- whether a system is at equilibrium by measuring the concentrations of the reactants and products.
- which direction a reaction will proceed by measuring the concentrations of reactants and products
- The extent of a reaction
- The pressure inside the container will not change
- The temperature will not change
- The temperature will increase
- the temperature will decrease
If for the formation of from and is 48 , then for decomposition of 1 mole of is________
What happens when pressure is increased in the following equilibrium rection
(i) N2O4(g)=2NO2(g)
(ii) H2(g)+I2(g)=2HI(g)
The ΔH∘f for CO2(g), CO(g) and H2O(g) are −393.5, −110.5 and −241.8kJ mole−1 respectively. The standard enthalpy change in kJ for the reaction CO2(g) + H2(g) → CO(g) + H2O(g) is
+524.1
+41.2
-262.5
-41.2
For a reaction ΔH = 9.08 kJ mol−1 and ΔS = 35.7 JK−1mol−1. Which of the following statements is correct for the reaction
Reversible and Isothermal
Reversible and Exothermic
Spontaneous and Endothermic
Spontaneous and Exothermic
Calculate △G⊖ for the following reaction:
CO(g)+12O2(g)→CO2(g); △H⊖ = -282.84 kJ
Given,
S⊖Co2=213.8J K−1mol−1, S⊖CO(g) = 197.9 J K−1mol−1,
S⊖Co2=205.0J K−1mol−1
-357 KJ
-383 KJ
-257 KJ
+ 357 KJ
A reversible chemical reaction having two reactants in equilibrium. If the
concentrations of the reactants are doubled, then the equilibrium constant
will
Also be doubled
Be halved
Become one-fourth
Remain the same
In the following reversible reaction 2SO2+O2 ⇌ 2SO3+ Q Cal
Most suitable condition for the higher production of SO3 is
High temperature and high pressure
High temperature and low pressure
Low temperature and high pressure
Low temperature and low pressure
Here,
R is gas constant :
- 0.5R
- R
- 1.5R
- 2R
- the forward reaction is exothermic
- the forward reaction is endothermic
- the slope of line is proportional to ΔH
- adding 'A' favours forward reaction
- removing C favours forward reaction
N2 (g)+3H2 (g)⇌2NH3 (g)
Given that ,
Keq=7×105
ΔS∘NH3=−23.66 cal mol−1K−1
Both the values are at 300K
Find the value of ΔH∘NH3 at 300 K
Take ln(7)=1.95
- −15.16 kcal mol−1
- −35.46 kcal mol−1
- −85.2 kcal mol−1
- −65.4 kcal mol−1
- Adding more oxygen
- Increasing the temperature at constant pressure
- Increasing the pressure at constant temperature
- Adding catalyst
For a given reversible reaction if the concentrations of the reactants are quadrupled, the equilibrium constant will be
be one fourth of the previous value
be quadrupled
remain the same
be halved
where Q is negative. Most suitable condition for a low production of SO3 is:
- High temperature and high pressure
- High temperature and low pressure
- Low temperature and high pressure
- Low temperature and low pressure
For the chemical equilibrium, CaCO3(s) ⇌ CaO(s) + CO2(g), ΔH., can be determined from which one of the following plots
For an equilibrium reaction A (g) + B (g) ⇌ C (g) + D (g), ΔH = -ve. An increase in temperature would cause
An increase in the value of Keq
A decrease in the value of Keq
No change in the value of Keq
A change in Kcq which cannot be qualitatively predicted