Question

1. The value of ΔHo for the reaction below is -228 kJ. Calculate the value of ΔHorxn (in kJ) when 2.59 moles of NaOH (s) is formed in the reaction?

2 Na2O2 (s) + 2 H2O (l) → 4 NaOH (s) + O2 (g)

2.The value of ΔHo for the reaction below is -721 kJ.

2 S (s) + 3 O2 (g) → 2 SO3 (s)

amount of heat released during the reaction of 6.64 grams of S (s) is _____kJ.

Answer 1

1)

2 Na2O2 (s) + 2 H2O (l) → 4 NaOH (s) + O2 (g)

From given reaction,

when 4 moles of NaOH is formed, ΔHorxn = -228 KJ

So,

ΔHorxn = (-228 KJ)* number of moles of NaOH / 4

ΔHorxn = (-228 KJ)* 2.59 / 4

ΔHorxn = -148 KJ

Answer: -148 KJ

2)

2 S (s) + 3 O2 (g) → 2 SO3 (s)

From given reaction,

when 2 moles of S reacts, ΔHorxn = -721 KJ

So,

ΔHorxn = (-721 KJ)* number of moles of S / 2

Here,

mass of S = 6.64 g

molar mass of S = 32 g/mol

so, number of mol of S = mass / molar mass

= 6.64 g/ 32 g/mol

= 0.2075 mol

ΔHorxn = (-721 KJ)* number of moles of S / 2

ΔHorxn = (-721 KJ)* 0.2075 / 2

ΔHorxn = -74.8 KJ

negative sign shown that heat has been released

heat released = 74.8 KJ

Answer: 74.8 KJ