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Consider the following thermochemical equations: \(\begin{array}{ll}\mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{HCHO}_{2}(l)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-411 \mathrm{~kJ} \\ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) & \Delta H^{\circ}=-128 \mathrm{~kJ} \\ \mathrm{HCHO}_{2}(l) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-33 \mathrm{~kJ}\end{array}\) Suppose the first equation is reversed and divided by 2 , The second and third equations are multiplied by \(\frac{1}{2},\) and then the three adjusted equations are added. What is the net reaction, and what is the value of \(\Delta H^{\circ}\) for the net reaction?

Short Answer

Expert verified
\(\Delta H^{\circ}\) for the net reaction \(\mathrm{H}_{2}(g) \longrightarrow \frac{1}{2}\mathrm{O}_{2}(g)\) is \(\frac{411}{2} - \frac{128}{2} - \frac{33}{2} = 125\) kJ.

Step by step solution

01

Reverse and halve the first equation

When reversing the first reaction, we invert the products and reactants, and the sign of \(\Delta H^{\circ}\) also changes. So, it becomes: \(\frac{1}{2}\mathrm{CH}_{3}\mathrm{OH}(l) + \frac{1}{2}\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \frac{1}{2}\mathrm{O}_{2}(g) + \frac{1}{2}\mathrm{HCHO}_{2}(l)\), with \(\Delta H^{\circ} = +\frac{411}{2}\mathrm{kJ}\).
02

Halve the second equation

Multiplying the second equation by \(\frac{1}{2}\) gives: \(\frac{1}{2}\mathrm{CO}(g) + \mathrm{H}_{2}(g) \longrightarrow \frac{1}{2}\mathrm{CH}_{3}\mathrm{OH}(l)\), with \(\Delta H^{\circ} = -\frac{128}{2}\mathrm{kJ}\).
03

Halve the third equation

Multiplying the third equation by \(\frac{1}{2}\) gives: \(\frac{1}{2}\mathrm{HCHO}_{2}(l) \longrightarrow \frac{1}{2}\mathrm{CO}(g) + \frac{1}{2}\mathrm{H}_{2}\mathrm{O}(l)\), with \(\Delta H^{\circ} = -\frac{33}{2}\mathrm{kJ}\).
04

Add the adjusted equations

When adding the adjusted equations, some substances will appear on both sides of the equation and thus can be canceled out. After canceling, the net result will be: \(\mathrm{H}_{2}(g) \longrightarrow \frac{1}{2}\mathrm{O}_{2}(g)\).
05

Calculate the net \(\Delta H^{\circ}\) for the reaction

Add up the \(\Delta H^{\circ}\) values from the adjusted equations: \(\Delta H^{\circ} = \frac{411}{2} - \frac{128}{2} - \frac{33}{2}\) kJ. Simplify to find the net \(\Delta H^{\circ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy Change
Enthalpy change is a crucial concept in thermodynamics that relates to the heat exchange in a system during a chemical reaction under constant pressure. It's represented by the symbol \( \Delta H^\circ \).

When a reaction releases heat into its surroundings, it is said to be exothermic, and \( \Delta H^\circ \) will have a negative value. This indicates that the energy of the products is lower than that of the reactants. Conversely, an endothermic reaction absorbs heat, which results in a positive \( \Delta H^\circ \). Here, the products have higher energy compared to the reactants.

Understanding enthalpy change is essential because it helps predict whether a reaction is likely to occur spontaneously. It also conveys how much energy is needed for a reaction to proceed, or how much might be released.
Chemical Reactions
Chemical reactions involve the breaking and forming of chemical bonds between atoms, leading to the transformation of reactants into products. They are described by chemical equations that detail the starting substances, the final substances, and sometimes the states of these substances (such as solid, liquid, gas).

During a chemical reaction, the conservation of mass principles states that atoms are neither created nor destroyed. This balancing act ensures that the same number of atoms of each element is present both before and after the reaction. Chemical reactions can be of various types, such as synthesis, decomposition, single replacement, or double replacement, depending on the nature of the interaction between reactants.
Stoichiometry
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. This discipline is founded on the law of conservation of mass and the concept of the mole, which is a measure of quantity in chemistry.

Using stoichiometry, chemists can calculate how much reactant is needed to produce a certain amount of product, or conversely, how much product can be produced from a given amount of reactant.

Ratio and Proportions

Stoichiometric coefficients in a balanced equation indicate the ratio in which substances react or form products. These coefficients are used to compute the moles of each substance involved.

Limiting Reagents and Yield

Additionally, stoichiometry is critical for identifying the limiting reagent in a reaction – the substance that will be exhausted first, thus determining the amount of product formed. It is also applied when calculating theoretical yield, the maximum quantity of product that could be generated from a given amount of reactant under perfect conditions.

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Most popular questions from this chapter

What fundamental fact about \(\Delta H\) makes Hess's law possible?

If a system does \(45 \mathrm{~J}\) of work and receives \(28 \mathrm{~J}\) of heat, what is the value of \(\Delta E\) for this change?

If the mass of a truck is doubled- for example, when it is loaded - by what factor does the kinetic energy of the truck increase? By what factor does the kinetic energy change if the mass is one-tenth of the original mass?

Toluene, \(\mathrm{C}_{7} \mathrm{H}_{8}\), is used in the manufacture of explosives such as TNT (trinitrotoluene). A \(1.500 \mathrm{~g}\) sample of liquid toluene was placed in a bomb calorimeter along with excess oxygen. When the combustion of the toluene was initiated, the temperature of the calorimeter rose from \(25.000^{\circ} \mathrm{C}\) to \(26.413^{\circ} \mathrm{C}\). The products of the combustion were \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l),\) and the heat capacity of the calorimeter was \(45.06 \mathrm{~kJ}^{\circ} \mathrm{C}^{-1}\) (a) Write the balanced chemical equation for the reaction in the calorimeter. (b) How many joules were liberated by the reaction? (c) How many joules would be liberated under similar conditions if 1.000 mol of toluene was burned?

Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 \(\left(\mathrm{CFCl}_{3}\right)\) and Freon-12 \(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right),\) which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the "ozone hole" that develops at the end of the Antarctic winter. In other parts of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earth's inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, \(\mathrm{O}_{3}\). Possible reactions are $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{O}_{2}(g)+\mathrm{ClO}(g) & \Delta H^{\circ}=-126 \mathrm{~kJ} \\ \mathrm{ClO}(g)+\mathrm{O}(g) \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) & \Delta H^{\circ}=-268 \mathrm{~kJ} \\ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g) & \Delta H^{\circ}=? \end{array} $$ The \(\mathrm{O}\) atoms in the second equation come from the breaking apart of \(\mathrm{O}_{2}\) molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of \(\Delta H^{\circ}\) (in kilojoules) for the last equation, the net reaction for the removal of \(\mathrm{O}_{3}\) from the atmosphere.

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