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The standard enthalpy change of combustion [to CO2(g)andH2O(l)at 250 Cof the organic liquid cyclohexane, C6H12(l), is -3731.7kJmol-1. Determine the ΔHf0of C6H12(l).

Short Answer

Expert verified

The standard enthalpy value of the reactant(ΔHf°)ofC6H12is-58.86kJ/mol

Step by step solution

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01

Given data

ΔH(Standard enthalpy change) value of Cyclohexane) C6H12(l) is -3731.7kJmol-1.

C6H10(l)+17/2O2(g)6CO2(g)+5H2O(l).

02

Concept of Enthalpy

Enthalpy change is amount of heat absorbed or evolved in a reaction when carried out at a constant pressure. It is generally calculated as the difference in energy required for bond breaking in a chemical reaction and energy gained in forming new bonds in a chemical reaction.

It is represented byΔH.

03

Calculate the value of ΔHf° of C6H12

The enthalpy change is calculated as:

ΔH°=nΔHfo(Products)-nΔHfo(Reactants) ...... (1)

Where, ΔHforepresent the standard enthalpy value.

localid="1663737494734" ΔHfoofO2is0.

ΔHfoofH2Ois-285.83kJmol-1.

ΔHfoofCO2is-393.51kJmol-1

ΔH°=-3731.7kJmol-1

.

The given expression is localid="1663738633260" C6H10(l)+17/2O2(g)6CO2(g)+5H2O(l).

Substitute these values in equation (1).

ΔH=6ΔHf0[CO2(g)]+5ΔHf0[H2O(l)]-{ΔHfa[C6H10(l)]+172×ΔHf0[O2(g)]}

-3731.7kJmol-1=6×-393.51kJmol-1+5×-285.83kJmol-1-{ΔHfa[C6H10(l)]+17/2×0]}ΔHfa[C6H10(l)]=-2361.06kJmol-1-1429.5kJmol-1+3731.7kJmol-1

Therefore,=-58.86k]mol.

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