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The equilibrium constant for the reaction, N2 (g)+O2 (g) = 2NO(g) is 4×10^−4 at 2000 k. In the presence of a catalyst, the equilibrium is attained ten times faster. Therefore, the equilibrium constant in the presence of the catalyst of 2000 k is:
11
Nov
The equilibrium constant for the reaction, N2 (g)+O2 (g) = 2NO(g) is 4×10^−4 at 2000 k. In the presence of a catalyst, the equilibrium is attained ten times faster. Therefore, the equilibrium constant in the presence of the catalyst of 2000 k is: N2 (g)+O2 (g) = 2NO(g) is 4×10^−4 at 2000 k. [...]
For the reaction, CO(g) + Cl2(g) ⇌ COCl2(g) ,the value of Kp /Kc is equal to:
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Nov
For the reaction, CO(g) + Cl2(g) ⇌ COCl2(g) ,the value of Kp /Kc is equal to: CO(g) + Cl2(g) ⇌ COCl2(g) For the reaction the value of Kp /Kc is equal to: November 11, 2020 Category: Chapter 5 - Chemical Equilibrium (Level 1 and Level 2) , N Awasthi Physical Chemistry ,
For the reversible reaction, N2(g)+3H2(g) ⇌ 2NH3(g) at 500 degree C, the value of Kp is 1.44×10^−5 when partial pressure is measured in the atmosphere. The corresponding value of Kc with concentration in mole litre ^−1 , is :
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Nov
For the reversible reaction, N2(g)+3H2(g) ⇌ 2NH3(g) at 500 degree C, the value of Kp is 1.44×10^−5 when partial pressure is measured in the atmosphere. The corresponding value of Kc with concentration in mole litre ^−1 , is : For the reversible reaction is N2(g)+3H2(g) ⇌ 2NH3(g) at 500 degree C the value of [...]
For the equilibrium SO2 Cl2(g) ⇌ SO2(g) + Cl2(g), what is the temperature at which kc(M) / kp (atm) = 3 ?
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Nov
For the equilibrium SO2 Cl2(g) ⇌ SO2(g) + Cl2(g), what is the temperature at which kc(M) / kp (atm) = 3 ? For the equilibrium SO2 Cl2(g) ⇌ SO2(g) + Cl2(g) what is the temperature at which kc(M) / kp (atm) = 3 ? November 11, 2020 Category: Chapter 5 - Chemical Equilibrium [...]
The equilibrium constant Kp for the following reaction at 191 degree C is 1.24 What is Kc ? B(s) + 2/3 F2(g) ⇌ BF3(g)
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Nov
The equilibrium constant Kp for the following reaction at 191 degree C is 1.24 What is Kc ? B(s) + 2/3 F2(g) ⇌ BF3(g) The equilibrium constant Kp for the following reaction at 191 degree C is 1.24 What is Kc ? B(s) + 2/3 F2(g) ⇌ BF3(g) November 11, 2020 Category: Chapter [...]
The equilibrium constant for the reaction N2(g)+O2(g)⇔2NO(g) at temperature T is 4×10^−4.The value of KC for the reaction, NO(g)⇔1/2N2(g)+1/2O2(g) at the same temperature is :
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Nov
The equilibrium constant for the reaction N2(g)+O2(g)⇔2NO(g) at temperature T is 4×10^−4.The value of KC for the reaction, NO(g)⇔1/2N2(g)+1/2O2(g) at the same temperature is : NO(g)⇔1/2N2(g)+1/2O2(g) at the same temperature is : The equilibrium constant for the reaction N2(g)+O2(g)⇔2NO(g) at temperature T is 4×10^−4.The value of KC for the reaction November 11, 2020 Category: Chapter [...]
The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are
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Nov
The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are and valency of an element with atomic number 15 are number of valence electrons respectively The group number November 11, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
The element having greatest difference between its first and second ionization energies , is (a) Ca (b) K (c) Ba (d) Sc
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Nov
The element having greatest difference between its first and second ionization energies , is (a) Ca (b) K (c) Ba (d) Sc is (a) Ca (b) K (c) Ba (d) Sc The element having greatest difference between its first and second ionization energies November 11, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + [...]
The IUPAC symbol for the element with atomic number 119 would be (a) unh (b) uun (c) une (d) uue
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Nov
The IUPAC symbol for the element with atomic number 119 would be (a) unh (b) uun (c) une (d) uue The IUPAC symbol for the element with atomic number 119 would be (a) unh (b) uun (c) une (d) uue November 11, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
The correct order of atomic radii is (a) Ce > Eu > Ho > N (b) N > Ce > Eu > Ho (c) Eu > Ce > Ho > N (d) Ho > N > Eu > Ce
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Nov
The correct order of atomic radii is (a) Ce > Eu > Ho > N (b) N > Ce > Eu > Ho (c) Eu > Ce > Ho > N (d) Ho > N > Eu > Ce The correct order of atomic radii is (a) Ce > Eu > Ho > N (b) [...]