Question

How much heat is released when 6.38 grams of Ag(s) (m.m = 107.9 g/mol) reacts by the equation shown below at standard state conditions?

4Ag_(s) + 2H₂S_(g) + O_2(g)     2Ag₂S_(s)+ 2H₂O_(l)

Substance         ∆Hºf (kJ/mol)

H₂S_(g)              -20.6

Ag₂S_(s)             -32.6

H₂O_(l) -285.8

Answer 1

Calculate ∆Hº reaction as shown below:

∆Hºrxn = ∆Hºf(products)-∆Hºf(reactants)

= [2mol∆Hºf(Ag₂S)+2mol∆Hºf(H₂O)] - [4mol∆Hºf(Ag) + 2mol∆Hºf(H₂S) + 1 mol ∆Hºf(O₂)]

= [2 mol(-32.6 kJ/mol)+2 mol(-285.8 kJ/mol)] - [4 mol(0.00 kJ/mol) + 2 mol(-20.6 kJ/mol) + 1 mol (0.00 kJ/mol)]

= [(-65.2 kJ) + (-571.6 kJ)] - [(-41.2 kJ)]

= -595.6 kJ

So, enthalpy change of -595.6 kJ occurs when 4 mol Ag reacts by given reaction.

Calculate the moles of Ag in 6.38 g as shown below:

Moles = mass/molar mass

= 6.38 g/107.9 g/mol

= 0.0591 mol

Now, enthalpy change when 0.0591 mol of Ag reacts by given reaction = (-595.6 kJ/4 mol)*0.0591 mol

= -8.80 kJ

As sign is negative, it means heat is released.

Hence, the heat released by 6.38 g of Ag will be -8.80 kJ. (Answer)