Chemistry, asked by meghakatiyar1, 1 year ago

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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Answered by Anonymous

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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 Anonymous

Answer:

2C6H6+15O2-------->12CO2+6H2O

n=12-15=-3

H=️E+️nRT

️H-️E=nRT

=>-3x8. 314x10^-3x298

= - 7.4kJ

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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.​

Answers

Hello mates

Follow me❤️

2C6H6+15O2-------->12CO2+6H2O

⚠️n=12-15=-3

⚠️H=⚠️E+⚠️nRT

⚠️H-⚠️E=⚠️nRT

=>-3x8. 314x10^-3x298

= - 7.4kJ

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.