Calculate the standard enthalpy of formation of liquid methanol, ch3oh(l), using the following information: c(graphite) + o2 --> co2(g) latex: \deltaδh° = –393.5 kj/mol h2(g) + (1/2)o2 --> h2o(l) latex: \deltaδh° = –285.8 kj/mol ch3oh(l) + (3/2)o2(g) --> co2(g) + 2h2o(l) latex: \deltaδh° = –726.4 kj/mol

Answer : The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation :

Standard formation of reaction : It is a chemical reaction that forms one mole of a substance from its constituent elements in their standard states.

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of will be,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

(3)    

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, we get :

(1)    

(2)    

(3)    

The expression for enthalpy of formation of will be,

Therefore, the standard enthalpy of formation of methanol is, -238.7 kJ/mole

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !

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