Question

If 0.685 mol of NH3(g) reacts with excess HCl(g) at 50°C for the following reaction, NH3(g) + HCl(g) → NH4Cl(s) what is the work done in terms of the system?

Answer 1

Answer :-
W = 1838.34(J) = 1.83834(kJ)

The balanced reaction between NH_3(g) and HCl_(g) is

NH3(g) + HCl) + NH4C)

Workdone in chemical reaction is given by

W = -RTAN

i.e.
W = -RT(n2-n1)
....(1)

Where,

W = workdone

R = gas constant

T = temperature

n₁ = number of moles of reactant

n₂ = number of moles of product

The given reaction is

NH3(g) + HCl) + NH4C)
...(A)

from reaction (A) it is observed that, NH₃, HCl and NH₄Cl having same moles ie. 0.685 mol each

From reaction we have,

n₁ = (0.685 +0.685) = 1.37 mol

n₂ = 0.685 mol

Again, T = 50 °C = (50 + 273) = 323 K

R = 0.082 L.atm/mol.K

From equation (1) we have,

W = -0.082(L.atm/mol.K) x 323(K)(0.685(mol) - (1.37(mol)

W = -0.082(L.atm/mol.K) x 323(K) x (-0.685(mol))

  

W = 18.143 L.atm)

We know

1 L.atm = 101.325 J

therefore,

18.143 L.atm =
18.143(L.atm) x 101.325(J) 2 = 1838.34(J) 1(L.atm)

therefore,

W = 1838.34(J) = 1.83834(kJ)

since, W = +Ve , work is on the system.