Question

physical chemistry, please show all work and explanations clearly

2. For the gas-phase reaction 2 NH3 <=> N2 + 3 H2, a closed system initially contained 1.00 mol of NH3, 4.50 mol of N2, and 4.20 mol of H2. Once equilibrium is achieved a) what are the minimum and maximum possible values for £? [-1.40 s E s 0.50] b) what are the minimum and maximum possible values for nNH,, NN,, and nu,?

Answer 1

The Greek symbol $\xi$ denotes the extent of reaction and is denoted by

Nin,i – neq, =

Here, n_eq,i demotes equilibrium concentration of i-th reactant and n_in,i denotes the initial concentration. ${\nu_i}$ here is the stoichiometric coefficient.

a) Maximum value of $\xi$ will be achieved when all of NH₃ will be converted to the reactant and n_eq for NH₃ becomes 0. n_in for NH₃ will be 1 mole and ${\nu_i}$ i.e. the stoichiometric coefficient is 2. Thus, we have,

1-0 Smar = =0.5

For minimum value of $\xi$ , the reaction should happen in reverse and maximum possible amount of NH₃ be formed. For the reverse reaction, 1 mole of N₂ will need 3 moles of H₂ and 4.50 moles of N₂ will need 3*4.50 = 13.50 moles of H₂ but there are only 4.20 moles. Hence, for the reverse reaction, H₂ will be the limiting reagent and it will go down to 0.

3 moles H₂ will form 2 moles NH₃. Thus, 4.20 moles of H₂ will form 4.20*2/3=2.80 moles NH₃. But 1.00 mole NH₃ was already there in the system.

Thus, for minimum $\xi$ , we have, n_eq for NH₃ will be 2.80+1.0 =3.80.

1-3.8 Smin = = -1.4

Thus, finally,

-1.40 < <0.5

b)

Here, we need to calculate maximum and minimum moles of each component.

Maximum extent of forward reaction will give minimum moles of NH₃ and maximum moles for H₂ and N₂

For NH₃,

піn,i — пеgi 1.00 – пеа

Putting , we will have,

{= Smar = 0.50

1.00 – neg 0.50 = neg=0.00

Putting , we will have,

<= min = -1.40

1.00 - ne -1.40 = = neg = 3.80

Thus, minimum possible value of n_NH3 is 0.00 moles and maximum possible value is 3.80 moles.

For N₂.

Here, n_in = 4.50 moles and v_i =1

Here, because Nitrogen is on product side, we will use the add a negative sign.

* = Mingi – Negi – 4.50 – neq

Putting , we will have,

{= Smar = 0.50

4.50 - ne 0.50 = - 1 neg= 5.00

Putting , we will have,

<= min = -1.40

-1.40 = _ 4.50 – neg neg= 3.10

Thus, minimum possible value of n_N2 is 3.10 moles and maximum possible value is 5.00 moles.

For H₂

Here, n_in = 4.20 moles and v_i =3

Here, because Hydrogen is on product side, we will use the add a negative sign.

піn,i — пеgi 4.20 – пеа S = --

Putting , we will have,

{= Smar = 0.50

4.20 – ne 0.50 = = neg= 5.70

Putting , we will have,

<= min = -1.40

4.20 - neg -1.40 = - = neg = 0.00

Thus, minimum possible value of n_H2 is 0.00 moles and maximum possible value is 5.70 moles.