Answer:
12.78 kJ
Explanation:
The correct balanced reaction would be
[tex]2CH_3OH\rightarrow 2CH_4+O_2\Delta H=252.8\ \text{kJ}[/tex]
Mass of methanol = [tex]3.24\ \text{g}[/tex]
Moles of methanol can be obtained by dividing the mass of methanol with its molar mass [tex](32.04\ \text{g/mol})[/tex]
[tex]\dfrac{3.24}{32.04}=0.10112\ \text{moles}[/tex]
Enthalpy change for the number of moles is given by
[tex]\dfrac{\text{Number of moles of methanol in the reaction}}{\text{Enthalpy change in the reaction}}=\dfrac{\text{Number of moles in 3.24 g of methanol}}{\text{Enthaply in change in the mass of methanol}}[/tex]
[tex]\\\Rightarrow\dfrac{2}{252.8}=\dfrac{0.10112}{\Delta H}\\\Rightarrow \Delta H=\dfrac{0.10112\times 252.8}{2}\\\Rightarrow \Delta H=12.781568\approx 12.78\ \text{kJ}[/tex]
The change in enthalpy is 12.78 kJ.
When 3.24 g of CH₃OH are decomposed, the change in the enthalpy is 12.8 kJ.
Let's consider the following thermochemical equation.
2 CH₃OH ⇒ 4 CH₄ + O₂ ∆H = 252.8 kJ
First, we will convert 3.24 g of CH₃OH to moles using its molar mass (32.04 g/mol).
[tex]3.24 g \times \frac{1mol}{32.04g} = 0.101 mol[/tex]
According to the thermochemical equation, 252.8 kJ are absorbed when 2 moles of CH₃OH react. The enthalpy change when 0.101 moles of CH₃OH react is:
[tex]0.101 mol \times \frac{252.8 kJ}{2mol} = 12.8 kJ[/tex]
When 3.24 g of CH₃OH are decomposed, the change in the enthalpy is 12.8 kJ.
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