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Calculate the standard enthalpy of forma...

Calculate the standard enthalpy of formation of `CH_(3)OH(l)` from the following data:
`CH_(3)OH(l)+3/2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l), …(i), Delta_(r)H_(1)^(Θ)=-726 kJ mol^(-1)`
`C(g)+O_(2)(g) rarr CO_(2)(g), …(ii), Delta_(c )H_(2)^(Θ)=-393 kJ mol^(-1)`
`H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l), ...(iii), Delta_(f)H_(3)^(Θ)=-286 kJ mol^(-1)` (Write magnitude of answer and in KJ / mol value)

Text Solution

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The correct Answer is:
293
The required equation for the formation of `CH_(3)OH(l)` is :
`C(s) + 2H_(2)(g) + 1/2O_(2)(g) to CH_(3)OH(l)`
The given equation are :
(i) `CH_(3)OH(l) + 3/2 O_(2)(g) to CO_(2)(g) + 2H_(2)O(l) Delta_(r)H^(@) = -726 kJ mol^(-1)`
`C(s)+ O_(2)(g) to CO_(2)(g) Delta_(r)H^(@) = -393 kJ mol^(-1)`
(iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) Delta_(r)H^(@) = -286.0 kJ mol^(-1)`
Multiply eq.(iii) by 2 and add to eq.(ii)
(ii) `C(s) + O_(2)(g) to CO_(2)(g) Delta_(r)` "`H^(@) = -393 kJ mol^(-1)`
`(iii) xx 2` "`2H_(2)(g) + O_(2)(g) to 2H_(2)O(l)` "`Delta_(r)H^(@) = -572 kJ mol^(-1)`
Adding (i) and (iii) we get
(iv) `C(s) + 2H_()(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l) Delta_(r)H^(@) = -965 kJ mol^(-1)`
Subtract eq.(i) from eq(iv)
(i) `CH_(3)OH(l) + 3/2 O_(2)(g) to CO_(2)+ 2H_(2)O(l) Delta_(r)H^(@) = -726 kJ mol^(-1)`
`C(s) + 2H_(2)(g) + 1/2 O_(2)(g) to CH_(3)OH(l)`
`Delta_(r)H^(@) = -239 kJ mol^(-1)`
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