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Calculate the Standard Enthalpy of Formation of Ch3oh(L) From the Following Data: - Chemistry

Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

CH3OH(l) + `3/2` O2(g) →CO2(g) + 2H2O(l) ; ΔrHθ = –726 kJ mol–1

C(g) + O2(g) →CO2(g) ; ΔcHθ = –393 kJ mol–1

H2(g) +`1/2` O2(g) → H2O(l) ; ΔfHθ = –286 kJ mol–1.

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Solution 1

The reaction that takes place during the formation of CH3OH(l) can be written as:

`C_(s) + 2H_2O_(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

Solution 2

The equation we aim at;

C(s) + 2H2(g) + l/202(g) ———> CH3OH (l);∆fH = ±? … (iv)

Multiply eqn. (iii) by 2 and add to eqn. (ii)

C(s) + 2H2(g) + 202(g) ————->C02(g) + 2H20(Z)

∆H = – (393 + 522) = – 965 kj moH Subtract eqn. (iv) from eqn. (i)

CH3OH(Z) + 3/202(g) ————> C02(y) + 2H20(Z); ∆H = – 726 kj mol-1

Subtract: C(s) + 2H2(y) + l/202(g) ———-> CH3OH(Z); ∆fHe = – 239 kj mol-1

Concept: Enthalpy Change, ∆_rH of a Reaction - Reaction Enthalpy - Standard Enthalpy of Reactions
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 6 Thermodynamics
Q 14 | Page 183
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