Question

1. the combustion of methanol is shown by the following equation:

2 ch₃oh (l)+3 o₂(g)→2 co₂(g)+4h₂o (g)
given the data below, use hess’s law to find the enthalpy change of the reaction for the equation above, then state the molar heat of combustion of methanol based on the equation formed.
c (s)+o₂(g)→co₂(g)+393 kj
h₂(g)+½ o₂(g)→h₂o (g)+242 kj
c (s)+2 h₂(g)+½ o₂(g)→ch₃oh (l)+638 kj

Explanation:

Step1: Manipulate the given equations

• Multiply the first equation \(C_{(s)}+O_{2(g)}\to CO_{2(g)}+393\ kJ\) by \(2\): \(2C_{(s)} + 2O_{2(g)}\to2CO_{2(g)}+786\ kJ\)
• Multiply the second equation \(H_{2(g)}+\frac{1}{2}O_{2(g)}\to H_{2}O_{(g)}+242\ kJ\) by \(4\): \(4H_{2(g)}+2O_{2(g)}\to4H_{2}O_{(g)}+968\ kJ\)
• Reverse the third equation \(C_{(s)}+2H_{2(g)}+\frac{1}{2}O_{2(g)}\to CH_{3}OH_{(l)}+638\ kJ\) to get \(CH_{3}OH_{(l)}\to C_{(s)}+2H_{2(g)}+\frac{1}{2}O_{2(g)}- 638\ kJ\), and then multiply by \(2\): \(2CH_{3}OH_{(l)}\to2C_{(s)}+4H_{2(g)}+O_{2(g)}-1276\ kJ\)

Step2: Add the manipulated equations

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After canceling out the common terms (\(2C_{(s)}\), \(4H_{2(g)}\)), we get \(2CH_{3}OH_{(l)}+3O_{2(g)}\to2CO_{2(g)}+4H_{2}O_{(g)}+(786 + 968-1276)\ kJ\)

Step3: Calculate the enthalpy change

\(\Delta H=(786 + 968-1276)\ kJ=478\ kJ\). Since the reaction is exothermic (heat is released), \(\Delta H=- 478\ kJ\)

Step4: Find the molar heat of combustion

For the equation \(2CH_{3}OH_{(l)}+3O_{2(g)}\to2CO_{2(g)}+4H_{2}O_{(g)}\), if \(2\) moles of \(CH_{3}OH\) release \(478\ kJ\) of heat. Then for \(1\) mole of \(CH_{3}OH\), the molar heat of combustion \(q=\frac{-478}{2}=-239\ kJ/mol\)

Answer:

The enthalpy change of the reaction is \(-478\ kJ\) and the molar heat of combustion of methanol is \(-239\ kJ/mol\)