A gaseous species A decomposes into gaseous species B and C as follows 2 A gas 2

Question

A gaseous species A decomposes into gaseous species B and C as follows

2 Agas 2Bgas + Cgas

The dissociated proportion of A increases with the temperature. 0.4 mol of A at 20°C is put in a reactor of 6 L that was emptied down to vacuum before. Then, it is heated. At 482 °F, the pressure in the reactor is 3.3 atm.

1. Show that this observation proves the existence of a chemical reaction.

2. What is, at 482 °F, the proportion (in %) of A that has been dissociated?

3. What are the partial pressures of the different gases in the mixture at 482 °F?

4. At 1472 °F , the dissociation of A is total. Calculate the pressure in the reactor.

5. Which would be the pressure at 1000 °C?

Explanation / Answer

Ans.1. A -------> 2 B + C

Stoichiometry: 1 mol reactant (A) produces 3 moles of product upon decomposition.

Given,

            Initial moles of A = 0.4 mol

            Temperature = 200C = 293 K

            Volume of reactor = 6.0 L

            Reactor is evacuated before A is introduced. So, the pressure in the vessel instantly after adding A (decomposition has not yet started) is solely due to presence of A.

            The vessel is heated to 4820F = 523.15 K

            Pressure at 523.15 K = 3.3 atm

Now,

Using Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K)

Pressure at 523.15 K if no A were decomposed is-

            P x 6.0 L = 0.4 mol x (0.0821 atm L mol-1K-1) x 523.15 K

            Or, P = 2.86 atm.

Thus, if there was no decomposition of A, the initial pressure (0.4 mol A, 523.15 K, 6.0 L vessel) should have been 2.86 atm.

Increase in pressure indicates that some A has decomposed into B and C. Since, 1 mol A decomposes into 3 mol products, relative increase in moles after decomposition causes increase in pressure.

So, increase in pressure from 2.86 atm to 3.3 atm indicates that decomposing of A has occurred.

Ans. 2. Let’ number of moles of A decomposed = X

Remaining moles of A = (0.4 -X)

Moles of B formed = 2 x moles of A decomposed = 2X mol

Moles of C formed = moles of A decomposed = X mol

Total number of moles = Remaining moles of A + moles of B formed + moles of C formed

                                    = (0.4-X) + 2X + X = (0.4 + 2X) mol ---- statement 1

Now, Given pressure of the vessel = 3.3 atm

Putting the values in equation 1-

            3.3 atm x 6.0 L = n x (0.0821 atm L mol-1K-1) x 523.15 K

            Hence, n = 0.461234 mol

Thus, total number of moles of all gases = 0.461234 mol         - statement 2

Now, comparing statement 1 and 2 –

            (0.4 + 2X) mol = 0.461234 mol

                Or, 2X = 0.461234 – 0.40 = 0.061234

                Hence, X = 0.030617 mol

That is, moles of A decomposed = 0.030617 mol

Thus, remaining moles of A = (0.4 -X ) mol = (0.4 - 0.030617) mol = 0.369383 mol

% A dissociated = (Moles of A dissociated / total initial moles of A) x 100

                                = (0.030617 mol / 0.4 mol) x 100

                                = 7.65 %

Ans. 3. At 523.15 K, when 0.030617 mol A has decomposed-

Total number of moles = 0.461234 mol            ; [see, Ans 2]

#3.A. Moles of A remaining = 0.369383 mol

            Moles fraction of A = remaining moles of A / total number of moles

                                                = 0.369383 mol/ 0.461234 mol = 0.800858133

            Partial pressure of A = moles fraction of A x Total pressure

                                                = 0.800858133 x 3.3 atm

                                                = 2.642 atm

#3.B. Moles of B Produced = 2 (X) = 2 x 0.030617 mol = 0.061234 mol

                Mole fraction of B = 0.061234 mol / 0.461234 mol = 0.1327

                Partial pressure of B= 0.1327 x 3.3 atm = 0.438 atm

#3.C. Moles of C Produced = (X) = 0.030617 mol

                Mole fraction of C = 0.030617 mol / 0.461234 mol = 0.066

                Partial pressure of C = 0.0666 x 3.3 atm = 0.219 atm

Ans. 4. 1 mol A produced 3 moles product upon decomposition.

So, total number of moles of product when decomposition is complete = 3 x Initial moles of A

                                                                                = 3 x 0.4 mol = 1.2 mol

Temperature = 14720F = 1073.15 K

Volume = 6.0 L

Putting the values in equation 1-

            P x 6.0 L = 1.2 mol x (0.0821 atm L mol-1K-1) x 1073.15 K

            Or, P = 17.61 atm

Thus, pressure of the vessel at 14720F = 17.61 atm.

Ans. 5. 10000C = 1273.15 K

n = 1.2 mol

Temperature = 1273.15 K

Volume = 6.0 L

Putting the values in equation 1-

            P x 6.0 L = 1.2 mol x (0.0821 atm L mol-1K-1) x 1273.15 K

            Or, P = 20.89

Hence, pressure at 10000C = 20.89 atm.

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