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प्रश्न
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
\[\ce{CH3OH_{(l)} + 3/2 O2_{(g)} →CO_{2(g)} + 2H2O_{(l)}}\] ; ΔrHθ = –726 kJ mol–1
\[\ce{C_{(g)} + O_{2(g)} →CO_{2(g)}}\] ; ΔcHθ = –393 kJ mol–1
\[\ce{H_{2(g)} +1/2 O_{2(g)} → H2O_{(l)}}\] ; ΔfHθ = –286 kJ mol–1.
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उत्तर
The reaction that takes place during the formation of CH3OH(l) can be written as:
\[\ce{C_{(s)} + 2H_2_{(g)} + 1/2 O_{2(g)} -> CH_3OH_{(l)}}\] .....(1)
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) – equation (i)
ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] – ΔrHθ
= (- 393 kJ mol-1) + 2(- 286 kJ mol-1) - (- 726 kJ mol-1)
= (- 393 - 572 + 726) kJ mol-1
∴ ΔfHθ [CH3OH(l)] = - 239 kJ mol-1
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