Question

The combustion of C2H4O2(l) has a delta H reaction value of -786.3 kJ. What is the enthalpy change when 180g of H2O(g) is produced? (Please include the sign)

Answer 1

The reaction is:

C2H4O2 + O2 —> CO2 + 2 H2O

Molar mass of H2O = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = 180 g

we have below equation to be used:

number of mol of H2O,

n = mass of H2O/molar mass of H2O

=(180.0 g)/(18.016 g/mol)

= 9.991 mol

Since delta H is negative, heat is released

when 2 mol of H2O is produced, heat released = 786.3 KJ

So,

for 9.991 mol of H2O, heat released = 9.991*786.3/2 KJ

= 3.928*10^3 KJ

Since this is heat is released, it is negative

Answer: - 3.928*10^3 KJ