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Question
Calculate the standard enthalpy of formation of liquid methanol from the following data:
- \[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\] ∆H° = – 726 kJ mol–1
- \[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\] ∆cH° = – 393 kJ mol–1
- \[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\] ∆fH° = – 286 kJ mol–1
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Solution
Given: Given equations are,
\[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\] ∆H° = – 726 kJ mol–1 ......(a)
\[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\] ∆cH° = – 393 kJ mol–1 .....(b)
\[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\] ∆fH° = – 286 kJ mol–1 .....(c)
To find: The standard enthalpy of formation (∆fH°) of CH3OH(l)
Calculation: Required equation is,
\[\ce{C_{(graphite)} + 2H_{2(g)} + \frac{1}{2} O_{2 (g)} -> CH3OH_{(l)}}\]
Multiply equation (c) by 2 and add to equation (b),
| \[\ce{2H_{2(g)} + O_{2(g)} -> 2H2O_{(l)}}\], | ∆rH° = – 572 kJ mol–1 |
| \[\ce{C_{(graphite)} + O_{2(g)} -> CO_{2(g)}}\], | ∆cH° = – 393 kJ mol–1 |
| \[\ce{C_{(graphite)} + 2H_{2(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\] | |
∆rH° = – 572 – 393 = – 965 kJ mol–1 ….(d)
Reverse equation (a) and add to equation (d),
| \[\ce{CO_{2(g)} + 2H2O_{(l)} -> CH3OH_{(l)} + \frac{3}{2} O_{2(g)}}\], | ∆rH° = 726 kJ mol–1 |
| \[\ce{C_{(graphite)} + 2H_{2(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\] | ∆rH° = – 965 kJ mol–1 |
| \[\ce{C_{(graphite)} + 2H_{2(g)} + \frac{1}{2} O_{2(g)} -> CH3OH_{(l)}}\] | |
∆fH° = ∆rH° = 726 – 965 = – 239 kJ mol–1
The standard enthalpy of formation (∆fH°) of CH3OH(l) from the given data is – 239 kJ mol–1
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