Consider the formation of hydrogen fluoride: H2(g) + F2(g) ↔ 2HF(g) If a 3.8 L nickel...

Consider the formation of hydrogen fluoride:

H₂(g) + F₂(g) ↔ 2HF(g)

If a 3.8 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0054 M H₂ is connected to a 3.6 L container filled with 0.036 M F₂. The equilibrium constant, K_p, is 7.8 x 10¹⁴ (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

Solutions

Expert Solution

Very large equilibrium constant means its product dominant reaction or reaction goes to completion .

After connection , Total volume = 3.8 + 3.6 = 7.4 litres.

Moles of Hydrogen = 3.8 x 0.0054 = 0.02052 moles

Moles of F2 = 3.6 x 0.036 = 0.1296 moles

New molarity after connecting ,

[H2] = 0.02052/7.4 = 0.00277 M limiting Reagent

[F2] = 0.1296/7.4 = 0.0175 M

H₂(g) + F₂(g) ↔ 2HF(g)

0.00277 0.0175 0

(0.00277-0.00277 ) (0.0175 - 0.00277) 2 x 0.00277= 5.54 x 10^-3 M

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

Consider the formation of hydrogen fluoride: H2(g) + F2(g) ↔ 2HF(g) If a 1.8 L nickel...

Consider the formation of hydrogen fluoride: H2(g) + F2(g) ↔ 2HF(g) If a 1.8 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0098 M H2 is connected to a 2.8 L container filled with 0.030 M F2. The equilibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?) Calculate the molar concentration of HF at equilibrium.

1)Consider the formation of hydrogen fluoride: H2(g) + F2(g) ? 2HF(g) If a 2.8 L nickel...

1)Consider the formation of hydrogen fluoride: H2(g) + F2(g) ? 2HF(g) If a 2.8 L nickel reaction container (glass cannot be used because it reacts with HF) filled with 0.0085 M H2 is connected to a 3.6 L container filled with 0.030 M F2. The equilibrium constant, Kp, is 7.8 x 1014 (Hint, this is a very large number, what does that imply?)Calculate the molar concentration of HF at equilibrium. 2)Suppose a 5.00 L nickel reaction container filled with 0.0052...

Consider the reaction H2 (g) + F2 (g) <-- --> 2HF (g) .At a certain temperature,...

Consider the reaction H2 (g) + F2 (g) <-- --> 2HF (g) .At a certain temperature, 0.50 moles of hydrogen gas, 0.50 moles of fluorine gas and .25 moles of hydrogen fluoride gas are placed in a 5.0 L container. Determine if the gas is at equilibrium or not and predict the direction it will move in order to get to equilibrium. And, for the mixure of the gases, determine the final concentrations of each gas. (use an appropriate table)....

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is...

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 64.0. A reaction mixture in a 10.00-L flask contains 0.23 moles each of hydrogen and fluorine gases plus 0.50 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium?

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is...

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 25.0. A reaction mixture in a 10.00-L flask contains 0.33 moles each of hydrogen and fluorine gases plus 0.30 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium?

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is...

At a particular temperature the equilibrium constant for the reaction: H2(g) + F2(g) ⇔ 2HF(g) is K = 81.0. A reaction mixture in a 10.00-L flask contains 0.25 moles each of hydrogen and fluorine gases plus 0.37 moles of HF. What will be the concentration of H2 when this mixture reaches equilibrium?

Arrange the following reaction in order of increasing S value. 1. H2(g) + F2(g) 2HF(g) 2....

Arrange the following reaction in order of increasing S value. 1. H2(g) + F2(g) 2HF(g) 2. NH4NO3(s) N2O(g) + 2H2O(I) 3. (NH4)2 Cr2O7(s) Cr2O3(s) + 4H2O(I) +N2(g)

The reaction of nitric oxide (NO(g)) with molecular hydrogen (H2(g)) results in the formation of nitrogen...

The reaction of nitric oxide (NO(g)) with molecular hydrogen (H2(g)) results in the formation of nitrogen and water as follows: 2NO (g) + 2H2(g)...> N2(g) + 2H2O(g) The experimentally determined rate-law expression for this reaction is first order in H2(g) and second order with NO(g). a) Is the reaction above, as written an elementary reaction? b) One potential mechanism for this reaction is as follows: H2(g) + 2NO(g)...> N2O (g) + H2O (g) k1 H2(g) + N2O (g)...> N2(g) +...

A.) For which reaction will Kp = Kc? 2 H2O(l) ↔ 2 H2(g) + O2(g) 2...

A.) For which reaction will Kp = Kc? 2 H2O(l) ↔ 2 H2(g) + O2(g) 2 HgO(s) ↔ Hg(l) + O2(g) S(s) + O2(g) ↔ SO2(g) H2CO3(s) ↔ H2O(l) + CO2(g) CaCO3(s) ↔ CaO(s) + CO2(g) B.) How does an increase in the pressure within the container by the addition of the inert gas E affect the amount of C present at equilibrium for the following reaction? A(g) + 3 B(g) ↔ 2 C(g) + D(g) ΔH° = -65 kJ...

1) Consider the equilibrium reaction N2(g) + 3 H2(g) ↔ 2 NH3(g) + 91.8 kJ Using...

1) Consider the equilibrium reaction N2(g) + 3 H2(g) ↔ 2 NH3(g) + 91.8 kJ Using Le Chatelier’s Principle, Name two specific ways this equilibrium can be shifted to the right to increase production of NH3. 2) In the reaction below, NH3(g) + H2O(l) ↔ NH2 - + H3O + The equilibrium constant is 1 x 10 -34. Is this reaction likely to take place? Explain your answer.

Subjects