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Question
Calculate the standard enthalpy of combustion of CH3COOH(l) from the following data:
`Delta_fH^@(CO_2)=-393.3 kJ mol^-1`
`Delta_fH^@(H_2O)=-285.8 kJ mol^-1`
`Delta_fH^@(CH_3COOH)=-483.2 kJ mol^-1`
Solution
`C_((s))+O_(2(g))->CO_(2(g))` `Delta_fH^@=-393.3 kJ mol^-1`...(1)
`H_(2(g))+1/2O_(2(g))->H_2O_((l))` `Delta_fH^@=-285.8 kJ mol^-1`...(2)
`2C_((s))+2H_(2(g))+O_(2(g))->CH_3COOH` `Delta_fH^@=-483.2 kJ mol^-1` ...(3)
The required equation is,
CH3COOH(l) + 2O2(g) →2CO2(g) + 2H2O(l)
Multiplying equation (1) and equation (2) by 2, then adding to reverse of equation (3).
`2C_((s))+2O_(2(g))->2CO_(2(g))` `Delta_fH^@=-786.6 kJ mol^-1`
`2H_(2(g))+1O_(2(g))->2H_2O_((l))` `Delta_fH^@=-571.6 kJ mol^-1`
`CH_3COOH_(l)->2C_((s))+2H_(2(g))+O_(2(g))` `Delta_fH^@= 483.2 kJ mol^-1`
--------------------------------------------------------------------------------------------
`CH_3COOH_(l)+2O_(2(g))->2CO_(2(g))+2H_2O_((l))` `Delta_fH^@=-875 kJ mol^-1`
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