Question of 98
Q.a) State Hess' law of constant heat summation.
(1)
b) Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
(3)
CH3OH(l) + 3/2 O2(g) → CO2(g) + H2O(l); ΔrH⊖ = –726 kJ mol-1
C (graphite) + O2(g) → CO2(g); ΔrH⊖ = –393 kJ mol-1
H2(g) + 1/2 O2(g) → H2O(l); ΔrH⊖ = –286 kJ mol-1
CH3OH(l) + 3/2 O2(g) → CO2(g) + H2O(l); ΔrH⊖ = –726 kJ mol-1
C (graphite) + O2(g) → CO2(g); ΔrH⊖ = –393 kJ mol-1
H2(g) + 1/2 O2(g) → H2O(l); ΔrH⊖ = –286 kJ mol-1
Kerala DhseKerala DHSE Plus One Board 2019Subjective· 4mImportance★★★★★
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Start your 14-day free trial to unlock the full solution →Hess's law says total enthalpy change of a reaction is the same however many steps it's split into, since enthalpy is a state function. Using it, combine the combustion data given (reversed) with the formation data for CO2 and H2O to build up methanol's own formation reaction, giving ΔfH⊖(CH3OH, l) = –239 kJ mol⁻¹.
- Hess's Law of Constant Heat Summation Hess's law states that if a chemical reaction is carried out in a series of steps, the standard enthalpy change for the overall (net) reaction is equal to the sum of the standard enthalpy changes of the individual steps, irrespective of the path or number of steps by which the reaction is brought about. This follows directly from the fact that enthalpy (H) is a state function — its change depends only on the initial and final states of the system, not on the route taken between them. This law lets us calculate the enthalpy of a reaction that is difficult (or impossible) to measure directly, by combining the enthalpies of other reactions that ARE measurable and that add up (algebraically) to the target reaction.
- Standard enthalpy of formation of CH3OH(l)
The target (formation) reaction we want ΔfH⊖ for is:
C(graphite) + 2H2(g) + 1/2 O2(g) → CH3OH(l) ... ΔfH⊖ = ?
We are given:
- CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l); ΔrH⊖ = –726 kJ mol⁻¹ (combustion of methanol)
- C(graphite) + O2(g) → CO2(g); ΔrH⊖ = –393 kJ mol⁻¹ (formation of CO2)
- H2(g) + 1/2 O2(g) → H2O(l); ΔrH⊖ = –286 kJ mol⁻¹ (formation of H2O) …
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