Question

Q1.The table below shows data for the four hydrocarbons ethyne, propyne, propene and propane. AHc is the

standard enthalpy of combustion of these hydrocarbons.
Compound
Name
M
-AHC
/kJ mol
HCECH
ethyne
26
1300
HC-CCH,
propyne
40
1940
H.C=CHCH,
propene
42
2060
CH,CH,CH
propane
44
2220
The complete combustion of 2.0 g of one of the above hydrocarbons releases exactly 100 kJ of
heat energy
This hydrocarbon is
A
ethyne
B
propyne
C
propene
D
propane
(Total 1 mark)​

Answer 1

`\large \boxed{\text{A. Ethyne}}`

Step-by-step explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be "moles" of "kJ"

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 26

2H-C≡C-H + … ⟶ 1300 kJ + …

m/g: 2.0

1. Moles of HC≡CH

`\text{Moles of HC$\equiv$CH} = \text{2.0 g} * \frac{\text{1 mol}}{\text{26 g}} = (1)/(13)\text{ mol}`

2. Heat released

The molar ratio is 1300 kJ:1 mol HC≡CH

`\text{Heat} = (1)/(13) \text{ mol HC$\equiv$CH } * \frac{\text{1300 kJ}}{\text{1 mol HC$\equiv$CH}} = \textbf{100 kJ}\\\\\text{The reaction of 2.0 g of $\large \boxed{\textbf{HC$\equiv$CH}}$ produces 100 kJ.}`