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Chapter 6: Problem 24

Consider this reaction: $$ \begin{aligned} 2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow & 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \\ \Delta H &=-1452.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ What is the value of \(\Delta H\) if (a) the equation is multiplied throughout by \(2,\) (b) the direction of the reaction is reversed so that the products become the reactants and vice versa, (c) water vapor instead of liquid water is formed as the product?

Short Answer

Expert verified
(a) -2905.6 kJ/mol, (b) +1452.8 kJ/mol, (c) -1290 kJ/mol

Step by step solution

01

(a) Doubling the equation

When the entire equation is multiplied by 2, the enthalpy change will also double. This is due to the fact that \(\Delta H\) is an extensive property and depends on the amount of substance involved in the reaction. Hence, \(\Delta H\) for the new reaction will be -1452.8 kJ/mol x 2 = -2905.6 kJ/mol.
02

(b) Reversing the equation

The enthalpy change of a reaction is the difference in the enthalpy of the products and the reactants. If the reaction is reversed, the signs of \(\Delta H\) will change as well. Thus, when the direction of the reaction is reversed, \(\Delta H\) will be +1452.8 kJ/mol.
03

(c) Changing the state of the Product

When water is in gaseous form rather than the liquid form, there will be an additional enthalpy change due to the vapourisation of water. We need to add this to the original \(\Delta H\). The heat of vaporization of water is about 40.7 kJ/mol. But in the equation, 4 moles of water are produced. Therefore, an additional 4 x 40.7 kJ/mol = 162.8 kJ of energy is needed to convert them into steam. Thus, the new \(\Delta H\) will be -1452.8 kJ/mol + 162.8 kJ/mol = -1290 kJ/mol.

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