N Awasthi Physical Chemistry
Ammonium carbamate dissociates as NH2 COONH4(s) ⇋2NH 3(g) +CO 2(g) Ina closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of NH3 now equals to the original total pressure. calculate the ratio of total pressure of CO2 now to the original partial pressure CO2.
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Nov
Ammonium carbamate dissociates as NH2 COONH4(s) ⇋2NH 3(g) +CO 2(g) Ina closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of NH3 now equals to the original total pressure. calculate the ratio of total pressure of CO2 now to the original partial pressure CO2. ammonia is added such [...]
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ammonia is added such that the partial pressure of NH3 now equals to the original total pressure. calculate the ratio of total pressure of CO2 now to the original partial pressure CO2. ,
Ammonium carbamate dissociates as NH2 COONH4(s) ⇋2NH 3(g) +CO 2(g) Ina closed vessel containing ammonium carbamate in equilibrium ,
For the equilibrium: LiCl.3NH3(s)⇔LiCl.NH3(s)+2NH3, Kp = 9atm^2 at 37∘C. A 5 litre vessel contains 0.1 mole of LiCl.NH3. How many mole of NH3 should be added to the flask at this temperature to derive the backward reaction for completion?
11
Nov
For the equilibrium: LiCl.3NH3(s)⇔LiCl.NH3(s)+2NH3, Kp = 9atm^2 at 37∘C. A 5 litre vessel contains 0.1 mole of LiCl.NH3. How many mole of NH3 should be added to the flask at this temperature to derive the backward reaction for completion? For the equilibrium: LiCl.3NH3(s)⇔LiCl.NH3(s)+2NH3 Kp = 9atm^2 at 37∘C. A 5 litre vessel contains 0.1 mole [...]
Assume that the decomposition of HNO3 can be represented by the following equation 4HNO3(g)⇔4NO2(g)+2H2O(g)+O2(g) and the reaction approaches equilibrium at 400K temperature and 30 atm pressure. At equilibrium partial pressure of HNO3 is 2 atm Calculate Kc in (mol/L−K) at 400K (Use:R=0.08atm−L/mol−K)
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Nov
Assume that the decomposition of HNO3 can be represented by the following equation 4HNO3(g)⇔4NO2(g)+2H2O(g)+O2(g) and the reaction approaches equilibrium at 400K temperature and 30 atm pressure. At equilibrium partial pressure of HNO3 is 2 atm Calculate Kc in (mol/L−K) at 400K (Use:R=0.08atm−L/mol−K) The most stable oxides of nitrogen will be : November 11, 2020 Category: [...]
The gas A2 in the left flask allowed to react with gas B2 present in right flask as A2(g)+B2(g)⇔2AB(g),Kc=4 at 27^∘C. What is the concentration of AB when equilibrium is established ?
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Nov
The gas A2 in the left flask allowed to react with gas B2 present in right flask as A2(g)+B2(g)⇔2AB(g),Kc=4 at 27^∘C. What is the concentration of AB when equilibrium is established ? Kc=4 at 27^∘C. What is the concentration of AB when equilibrium is established ? The gas A2 in the left flask allowed to [...]
One mole of N2 (g) is mixed with 2 moles of H 2(g) in a 4 litre vessel. If 50% of N2(g) is converted to H2(g) by the following reaction : N2 (g)+3H2(g)⇌2NH3 (g) What will be the value of Kc for the following equilibrium? NH3 (g)⇌ 2/1 N2 (g) + 2/3 H2 (g)
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Nov
One mole of N2 (g) is mixed with 2 moles of H 2(g) in a 4 litre vessel. If 50% of N2(g) is converted to H2(g) by the following reaction : N2 (g)+3H2(g)⇌2NH3 (g) What will be the value of Kc for the following equilibrium? NH3 (g)⇌ 2/1 N2 (g) + 2/3 H2 (g) [...]
The following equilibrium constants were determined at 1120K: 2CO(g)⇔C(s)+CO2(g),,Kp1=10−114atm−1 CO(g)+Cl2(g)⇔COCl2(g),,Kp2=6×10−3atm−1 What is the equilibrium constant Kc for the following reaction at 1120K: C(s)+CO2(g)+2Cl2(g)⇔2COCl2(g)
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Nov
The following equilibrium constants were determined at 1120K: 2CO(g)⇔C(s)+CO2(g),,Kp1=10−114atm−1 CO(g)+Cl2(g)⇔COCl2(g),,Kp2=6×10−3atm−1 What is the equilibrium constant Kc for the following reaction at 1120K: C(s)+CO2(g)+2Cl2(g)⇔2COCl2(g) Kp1=10−114atm−1 CO(g)+Cl2(g)⇔COCl2(g) Kp2=6×10−3atm−1 What is the equilibrium constant Kc for the following reaction at 1120K: C(s)+CO2(g)+2Cl2(g)⇔2COCl2(g) The following equilibrium constants were determined at 1120K: 2CO(g)⇔C(s)+CO2(g) November 11, 2020 Category: Chapter 5 - [...]
The standard free energy change of a reaction is ΔG∘=−115 kJ mol^-1 at 298 K. Calculate the the value of log10 Kp (R=8.314 jK^−1mol^−1)
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Nov
The standard free energy change of a reaction is ΔG∘=−115 kJ mol^-1 at 298 K. Calculate the the value of log10 Kp (R=8.314 jK^−1mol^−1) The standard free energy change of a reaction is ΔG∘=−115 kJ mol^-1 at 298 K. Calculate the the value of log10 Kp (R=8.314 jK^−1mol^−1) November 11, 2020 Category: Chapter 5 - [...]
Solid Ca(HCO3) 2 decomposes as Ca(HCO3)2 (s)⇌CaCO3(s)+CO2 (g)+H2 O(g) If the total pressure is 0.2 bar at 420K what is the standard free energy change for given reaction (ΔrG^o )?
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Nov
Solid Ca(HCO3) 2 decomposes as Ca(HCO3)2 (s)⇌CaCO3(s)+CO2 (g)+H2 O(g) If the total pressure is 0.2 bar at 420K what is the standard free energy change for given reaction (ΔrG^o )? Solid Ca(HCO3) 2 decomposes as Ca(HCO3)2 (s)⇌CaCO3(s)+CO2 (g)+H2 O(g) If the total pressure is 0.2 bar at 420K what is the standard free [...]
For the reaction at 300K A(g)⇔V(g)+S(g) ΔrH∘=−30kJ/mol, ΔrS∘=−0.1Kj.k^−1.mol^−1 What is the value of equilibrium constant ?
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Nov
For the reaction at 300K A(g)⇔V(g)+S(g) ΔrH∘=−30kJ/mol, ΔrS∘=−0.1Kj.k^−1.mol^−1 What is the value of equilibrium constant ? For the reaction at 300K A(g)⇔V(g)+S(g) ΔrH∘=−30kJ/mol ΔrS∘=−0.1Kj.k^−1.mol^−1 What is the value of equilibrium constant ? November 11, 2020 Category: Chapter 5 - Chemical Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
A plot of Gibbs energy of a reaction mixture against the extent of the reaction is:
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A plot of Gibbs energy of a reaction mixture against the extent of the reaction is: A plot of Gibbs energy of a reaction mixture against the extent of the reaction is: November 11, 2020 Category: Chapter 5 - Chemical Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,