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The heat of combusion of benzene in a bo...

The heat of combusion of benzene in a bomb calorimeter (i.e constant volume) was found to be `3263.9kJ mo1^(-1)` at `25^(@)C` Calculate the heat of combustion of benzene at constan pressure .

Text Solution

Verified by Experts

The reaction is `C_(6)H_(6) (l) +7(1)/(2)O_(2)(g) rarr 76CO_(2)(g) + 3H_(2)O(l)`
In this reaction, `O_(2)` is the only gaseous reactant and `CO_(2)` is the only gaseous product.
`:. Delta N_(g)= n_(p) - n_(r) = 6-7(1)/(2) = -1(1)/(2)= - (3)/(2)`
Also, we are given `Delta U ( or q_(v))= - 3263.9 kJmol^(-1)`
`T= 25^(@)C = 298 K`
`R = 8.314 J K^(-1) mol^(-1) = (8.314)/(1000) kJ K^(-1) mol^(-1)`
`Delta H ( or q_(p))= Delta U + Delta n_(g) RT = - 3263.9- 3.7 kJ mol^(-1) + (-(3)/(2)mol) ((8.314)/(1000) kJ K^(-1) mol^(-1)) ( 298K)`
`= - 3263.9 - 3.7 kJ mol^(-1) = - 3267.6 kJ mol^(-1)`
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The heat produced by the combustion of 2.0g of benzene in a bomb calorimeter was found to be 123.6 kJ at 25^(@)C . Calculate the enthalpy of combustion of benzene at 25^(@)C .

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The combustion of benzene (l) gives CO_(2)(g) and H_(2)O(l) . Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol^(–1) at 25^(@)C , heat of combustion (in kJ mol^(–1) ) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1)

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Knowledge Check

  • The combustion of benzene (1) gives CO_(2) (g) and H_(2)O (l). Given that heat of combustion of benzene at constant volume is -3263.9kJ"mol"^(-1) at 25^(@)C , heat of combustion (in kJ"mol^(-1) ) of benzene at constant pressure will be ( R=8.314JK^(-1)"mol"^(-1) )

    A
    4152.6
    B
    `-452.46`
    C
    3260
    D
    `-3267.6`
  • The combustion of benzene (1) gives CO_(2)(g) and H_(2)O(1) . Given that heat of combustion of benzene at constant volume is .^(-)3263.9 kJ mol^(-1) " at "25^(@)C , heat of combustion (in kJ mol^(-1) ) of benzene at constant pressure will be (R=8.314 JK^(-1) mol^(-1))

    A
    `- 452.46`
    B
    3260
    C
    `- 3267.6`
    D
    4152.6
  • The combustion of benzene (l) gives CO_(2)(g) and H_(2)O(l) . Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol^(–1) at 25^(@)C , heat of combustion (in kJ mol^(–1) ) of benzene at constant pressure will be (R = 8.314 JK–1 mol–1)

    A
    `-3267.6`
    B
    `4152.6`
    C
    `-452.46`
    D
    3260