हिंदी

Calculate the standard enthalpy of formation of liquid methanol from the following data:

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प्रश्न

Calculate the standard enthalpy of formation of liquid methanol from the following data:

  1. \[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]     ∆H° = – 726 kJ mol–1
  2. \[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\]          ∆cH° = – 393 kJ mol–1
  3. \[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\]          ∆fH° = – 286 kJ mol–1 
संख्यात्मक
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उत्तर

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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