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Chapter 6: Problem 115

Consider the following equations: $$ \begin{array}{ll}{3 \mathrm{A}+6 \mathrm{B} \longrightarrow 3 \mathrm{D}} & {\Delta H=-403 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{E}+2 \mathrm{F} \longrightarrow \mathrm{A}} & {\Delta H=-105.2 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{C} \longrightarrow \mathrm{E}+3 \mathrm{D}} & {\Delta H=64.8 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6},\) the second and third equations are divided by \(2,\) and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

Short Answer

Expert verified
The net reaction is \(\frac{1}{2}C + F \longrightarrow \frac{1}{2}A + B + \frac{1}{2}D\) and the overall heat of this reaction is \(\Delta H_{total} = 46.97 \, \text{kJ/mol}\).

Step by step solution

01

Reverse and Multiply First Equation

We need to reverse the first equation, which means changing the reactants into products and vice versa. Then, multiply each coefficient by \(\frac{1}{6}\). The reaction becomes: $$ \frac{1}{6}(3D \longrightarrow 3A + 6B) $$ After simplifying, we get: $$ D \longrightarrow \frac{1}{2}A + B $$ 2. Divide the second and third equations by 2:
02

Divide Second Equation by 2

Divide the coefficients of the second equation by 2: $$ \frac{1}{2}(E + 2F \longrightarrow A) $$ After simplifying, we get: $$ \frac{1}{2}E + F \longrightarrow \frac{1}{2}A $$
03

Divide Third Equation by 2

Divide the coefficients of the third equation by 2: $$ \frac{1}{2}(C \longrightarrow E + 3D) $$ After simplifying, we get: $$ \frac{1}{2}C \longrightarrow \frac{1}{2}E + \frac{3}{2}D $$ 3. Calculate the enthalpy changes for the adjusted reactions:
04

Calculate Adjusted Enthalpy Changes

For each adjusted equation, calculate the enthalpy change by utilizing the same factor used for the coefficients: For the first equation, since it was reversed initially and then multiplied by \(\frac{1}{6}\), the enthalpy change is multiplied by \(-\frac{1}{6}\): $$ \Delta H_1 = -\frac{1}{6}(-403) = 67.17 \, \text{kJ/mol} $$ For the second and third equations, the coefficients were divided by 2, so we also divide the initial enthalpy changes by 2: $$ \Delta H_2 = \frac{1}{2}(-105.2) = -52.6 \, \text{kJ/mol} $$ $$ \Delta H_3 = \frac{1}{2}(64.8) = 32.4 \, \text{kJ/mol} $$ 4. Add together all adjusted equations:
05

Add Adjusted Equations

Add together the adjusted chemical equations: $$ (D \longrightarrow \frac{1}{2}A + B) + (\frac{1}{2}E + F \longrightarrow \frac{1}{2}A) + (\frac{1}{2}C \longrightarrow \frac{1}{2}E + \frac{3}{2}D) $$ By combining like terms, we find the net reaction: $$ \frac{1}{2}C + F \longrightarrow \frac{1}{2}A + B + \frac{1}{2}D $$ 5. Calculate the overall heat of the combined reaction:
06

Determine Overall Heat

Add the enthalpy changes for all adjusted equations: $$ \Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 $$ $$ \Delta H_{total} = 67.17 - 52.6 + 32.4 = 46.97 \, \text{kJ/mol} $$ Hence, the net reaction is: $$ \frac{1}{2}C + F \longrightarrow \frac{1}{2}A + B + \frac{1}{2}D $$ And the overall heat of this reaction is: $$ \Delta H_{total} = 46.97 \, \text{kJ/mol} $$

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Most popular questions from this chapter

Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ} $$ a. How much heat is evolved for the production of 1.00 mole of $\mathrm{H}_{2} \mathrm{O}(l) ?$ b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when 186 \(\mathrm{g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was $2.0 \times 10^{8} \mathrm{L}\( at 1.0 atm and \)25^{\circ} \mathrm{C} .$ How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Assuming gasoline is pure \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) predict the signs of \(q\) and \(w\) for the process of combusting gasoline into \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\)

In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.

Consider a balloon filled with helium at the following conditions. $$ \begin{array}{l}{313 \mathrm{g} \mathrm{He}} \\ {1.00 \mathrm{atm}} \\ {1910 . \mathrm{L}} \\ {\text { Molar Heat Capacity }=20.8 \mathrm{J} / \mathrm{C} \cdot \mathrm{mol}}\end{array} $$ The temperature of this balloon is decreased by \(41.6^{\circ} \mathrm{C}\) as the volume decreases to \(1643 \mathrm{L},\) with the pressure remaining constant. Determine \(q, w,\) and $\Delta E(\text { in } \mathrm{kJ} \text { ) for the compression of }$ the balloon.

Hydrogen gives off \(120 . \mathrm{J} / \mathrm{g}\) of energy when burned in oxygen, and methane gives off \(50 . \mathrm{J} / \mathrm{g}\) under the same circumstances. If a mixture of 5.0 \(\mathrm{g}\) hydrogen and \(10 . \mathrm{g}\) methane is burned, and the heat released is transferred to 50.0 \(\mathrm{g}\) water at \(25.0^{\circ} \mathrm{C},\) what final temperature will be reached by the water?
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