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Standard VI
Chemistry
Burning And Combustion
Question

The heat of combustion of benzene at 27∘C found by bomb calorimeter i.e. for the reaction C6H6(l)+712O2(g)→6CO2(g)+3H2O(l) is 780 K.Cal mol−1. The heat evolved on burning 39g of benzene in an open vessel will be
  1. 780.9 K.Cal
  2. 390 K.Cal
  3. 390.45 K.Cal
  4. 780 K.Cal

A
780.9 K.Cal
B
780 K.Cal
C
390 K.Cal
D
390.45 K.Cal
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Similar Questions
Q1
The heat of combustion of benzene at 27∘C found by bomb calorimeter i.e. for the reaction C6H6(l)+712O2(g)→6CO2(g)+3H2O(l) is 780 K.Cal mol−1. The heat evolved on burning 39g of benzene in an open vessel will be
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Q2
For combustion of 1 mole of benzene at 250C, the heat of reaction at constant pressure is - 780.9 kcal. What will be the heat of reaction at constant volume?
C6H6(I)+712O2(g)→6CO2(g)+3H2O(l)
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Q3
Benzene burns in oxygen according to the following reaction :
C6H6(l)+152O2(g)→3H2O(l)+6CO2(g)
If the standard enthalpies of formation of C6H6(l), H2O(l) and CO2(g) are 11.7, -68.1 and -94 kcal/mole, respectively, the amount of heat that will liberate burning 780 g benzene is:
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Q4
The heat evolved in the combustion of benzene is given by
C6H6(l)+712O2(g)→3H2O(l)+6CO2(g),ΔH=−781.0Kcalmol−1
When 156gofC6H6 is burnt in an open container, the amount of heat energy released will be (in kJ/mol)
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Q5
The heat of combustion of benzene in a bomb calorimeter (i.e., constant volume) was found to be 3263.9 kJmol−1 at 25oC. Calculate the heat of combustion of benzene at constant pressure.
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