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Chapter 6: Problem 24

Consider this reaction: $$\begin{array}{r}2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \\\\\Delta H=-1452.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ 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
The \(\Delta H\) of the reaction, (a) when multiplied by 2 is \(-2905.6 \, kJ/mol\), (b) when the direction is reversed is \(1452.8 \, kJ/mol\), and (c) when water vapor is formed instead of liquid water is \(-1290 \, kJ/mol\).

Step by step solution

01

Calculating \(\Delta H\) when equation is multiplied by 2

Applying the rule of enthalpy states that when the stoichiometric coefficients of a thermochemical equation are multiplied by a factor, the \(\Delta H\) of the reaction is also multiplied by the same factor. Therefore, if the equation is multiplied throughout by 2, \(\Delta H\) will also be doubled. Hence, \(\Delta H\) for the reaction when multiplied by 2 would be \(2*(-1452.8 \, kJ/mol) = -2905.6 \, kJ/mol\).
02

Calculating \(\Delta H\) when direction of reaction is reversed

In the case of reversing the reaction, the sign of \(\Delta H\) simply becomes the opposite. This is because reversing a reaction makes an exothermic reaction endothermic and vice versa. So if \(\Delta H\) for the initial reaction is \(-1452.8 \, kJ/mol\), then for the reversed reaction, it will be \(1452.8 \, kJ/mol\).
03

Calculating \(\Delta H\) when water is vapor instead of liquid

Changing the state of water from liquid to gas would increase the \(\Delta H\) value of the reaction due to the heat of vaporization of water, which is approximately \(40.7 \, kJ/mol\). Therefore, for every mole of water that changes phase, it will contribute an extra \(40.7 \, kJ\) to the \(\Delta H\) of the reaction. Since there are 4 moles of water in the reaction equation, this will contribute \(4*40.7 = 162.8 \, kJ/mol\) to \(\Delta H\), making the final \(\Delta H\) for the new reaction: \(-1452.8 \, kJ/mol + 162.8 \, kJ/mol = -1290 \, kJ/mol\).

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Most popular questions from this chapter

Calculate the work done (in joules) when 1.0 mole of water is frozen at \(0^{\circ} \mathrm{C}\) and 1.0 atm. The volumes of 1 mole of water and ice at \(0^{\circ} \mathrm{C}\) are \(0.0180 \mathrm{~L}\) and \(0.0196 \mathrm{~L},\) respectively.

At \(25^{\circ} \mathrm{C},\) the standard enthalpy of formation of \(\mathrm{HF}(a q)\) is given by \(-320.1 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{OH}^{-}(a q),\) it is \(-229.6 \mathrm{~kJ} / \mathrm{mol} ;\) of \(\mathrm{F}^{-}(a q),\) it is \(-329.1 \mathrm{~kJ} / \mathrm{mol} ;\) and of \(\mathrm{H}_{2} \mathrm{O}(l),\) it is \(-285.8 \mathrm{~kJ} / \mathrm{mol} .\) (a) Calculate the standard enthalpy of neutralization of \(\mathrm{HF}(a q)\) $$ \mathrm{HF}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{F}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (b) Using the value of \(-56.2 \mathrm{~kJ}\) as the standard enthalpy change for the reaction $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$calculate the standard enthalpy change for the reaction$$\mathrm{HF}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q) $$

Which of the following standard enthalpy of formation values is not zero at \(25^{\circ} \mathrm{C} ? \mathrm{Na}(s), \operatorname{Ne}(g)\) \(\mathrm{CH}_{4}(g), \mathrm{S}_{8}(s), \mathrm{Hg}(l), \mathrm{H}(g)\)

Calculate the standard enthalpy change for the reaction $$2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)$$ given that$$\begin{array}{l}2 \mathrm{Al}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-1669.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-822.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{MHCl}\) is mixed with an equal volume of \(0.431 M \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\), For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?
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