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The combustion of 1 mole of benzene take...

The combustion of 1 mole of benzene takes place at 298K and 1 atm. After combustion, `CO_(2(g)) and H_(2)O_((l))` are produced and 3267.0 KJ of heat is liberated. Calculate the standard enthalpy of formation, `Delta_(f)H^(0)` of benene. Standard enthalpies of formation of `CO_(2(g))andH_(2)O_((l))` are -393.5 KJ `mol^(-1)` and - 285.83 KJ `mol^(-1)` respectively

Text Solution

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Aim `: 6C(s) + 3H_(2)(g) rarr C_(6)H_(6)(l),DeltaH= ?`
Given `: (i) C_(6) H_(6)(l) + (15)/(2) O_(2)(g) rarr 6 CO_(2)(g) + 3H_(2)O(l), Delta H = - 3267.0 kJ mol^(-1)`
(ii) `C(s) + O_(2)(g) rarr CO_(2)(g) , DeltaH = -393.5 kJ mol^(-1)`
(iii) `H_(2)(g) + (1)/(2) O_(2)(g) rarr H_(2)O(l) , Delta H = -285.83 kJ mol^(_1)`
In order to get th required thermochemical equation, multiply Eq. (ii) 6 and Eq. (iii) by 3 and subtract Eq. (i) from their sum, i.e., operating 6 `xx` Eqn. (ii) `+ 3 xx `Eqn (iii) - Eqn (i), we get
`6 C(s) + 3H_(2) (g) rarr C_(6) H_(6)(l) , DeltaH = 6(-393.5) + 3(-285.83) - ( -3267.0)`
`= - 2361- 857.49 + 3267.0`
`= - 48.51kJ mol^(-1)`
Thus, the enthalpy of formation of benzene is `Delta _(f) H= - 48.51kJ mol^(-1)`
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The combusition of 1 mol of benzene (C_(6) H_(6)) takes place at 298 K and 1 bar pressure. After combustion, CO_(2) (g) and H_(2) O (1) are produced and 3267 kJ of heat is liberated. Calculate the standard enthaply of formation, Delta_(f) H^(@) of benzene. Standard enthapies of formation of CO_(2) (g) and H_(2) O (1) are - 393.5 kJ mol^(-1) and - 258.83 kJ mol^(-1) , respectively. Strategy : Apply Eq. the mathematical form of Hesis's law, to the combustion reaction of 1 mol of benzene. Remember Delta_(f) H^(@) for O_(2) (g) is zero by convention. We are give Delta_(1) H^(@) and Delta_(f) H^(@) values for all substance except C_(6) H_(6) (1) . We can solve for this unknown.

The combustion of 1 mol of benzene takes place at 298 K and 1 atm . After combustion, CO_(2)(g) and H_(2)O(l) are produced and 3267.0 kJ of heat is librated. Calculate the standard entalpy of formation, Delta_(f)H^(Θ) of benzene Given: Delta_(f)H^(Θ)CO_(2)(g) = -393.5 kJ mol^(-1) Delta_(f)H^(Θ)H_(2)O(l) = -285.83 kJ mol^(-1) .

Knowledge Check

  • The standard enthalpy of formation of octane (C_(8)H_(18)) is -250 KJ/mol. Calculate the enthalpy of formation of CO_(2)(g) and H_(2)O_(l) are -394 KJ/mol and -286 KJ/ mol respectively :

    A
    -5200 KJ/mol
    B
    -5726 KJ/mol
    C
    -5476 KJ/mol
    D
    -5310 KJ/mol

  • The standard enthalpies of formation of CO_(2)(g) and HCOOH (l) are -393.7 kJ"mol"^(-1) and -409.2 kJ"mol"^(-1) respectively. Which of the following statements are correct?

    A
    `-393.7 kJ "mol"^(-1)` is the enthalpy change for the reaction, `C(s) + O_(2)(g) to CO_(2)(g)`
    B
    The enthalpy change for the reaction, `CO_(2)(g) + H_(2)(g) to HCOOH(l)`, would be `-15.5 kJ"mol"^(-1)`.
    C
    The enthalpy change for the reaction, `H_(2)(g) + CO_(2)(g) to H_(2)O(l) + CO(g)`, is `-409.2 kJ"mol"^(-1)`
    D
    The final temperature achieved is 291.8 K

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