Calculate the difference of heat of reaction at constant pressure and that at constant volume for the combustion of two moles of liquid benzene at 25°C.
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2C6H6+15O2-------->12CO2+6H2O
⚠️n=12-15=-3
⚠️H=⚠️E+⚠️nRT
⚠️H-⚠️E=⚠️nRT
=>-3x8. 314x10^-3x298
= - 7.4kJ
Answered by
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Answer:
2C6H6+15O2-------->12CO2+6H2O
n=12-15=-3
H=️E+️nRT
️H-️E=nRT
=>-3x8. 314x10^-3x298
= - 7.4kJ
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