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Chapter 6: Problem 75

The standard enthalpy change \(\Delta H^{\circ}\) for the thermal decomposition of silver nitrate according to the following equation is \(+78.67 \mathrm{~kJ}\) : $$\mathrm{AgNO}_{3}(s) \longrightarrow \mathrm{AgNO}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g)$$ The standard enthalpy of formation of \(\mathrm{AgNO}_{3}(s)\) is \(-123.02 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the standard enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\)

Short Answer

Expert verified
The standard enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\) is +201.69 kJ/mol.

Step by step solution

01

Analyze the given data and setup the equation

First, remember that the equation for enthalpy change is \(\Delta H^{\circ} = H_{f, products} - H_{f, reactants}\). For the decomposition reaction: \(\mathrm{AgNO}_{3}(s) \rightarrow \mathrm{AgNO}_{2}(s)+\frac{1}{2}\mathrm{O}_{2}(g)\), the provided enthalpy change is +78.67 kJ. The enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\) is what needs to be found, and the enthalpy of formation of \(\mathrm{AgNO}_{3}(s)\) is given as -123.02 kJ/mol. Let's denote the enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\) as \(x\). Also note that for elemental oxygen, enthalpy of formation is 0. Substituting these values into the equation gives us: \[78.67 kJ = (x + 0) - (-123.02 kJ)\]
02

Solve for the unknown

To solve for \(x\), arrange the above equation to isolate \(x\). This gives us: \(x = 78.67 kJ + 123.02 kJ\)
03

Calculate the final result

Add the values to get the final result for \(x\), which equals to 201.69 kJ. Therefore, the standard enthalpy of formation of \(\mathrm{AgNO}_{2}(s)\) is +201.69 kJ/mol.

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Most popular questions from this chapter

Determine the amount of heat (in kJ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of \(\mathrm{NO}_{2}\) are produced according to the equation $$\begin{array}{r}2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\\\Delta H=-114.6 \mathrm{~kJ} / \mathrm{mol}\end{array}$$

A driver's manual states that the stopping distance quadruples as the speed doubles; that is, if it takes \(30 \mathrm{ft}\) to stop a car moving at \(25 \mathrm{mph}\) then it would take \(120 \mathrm{ft}\) to stop a car moving at \(50 \mathrm{mph}\). Justify this statement by using mechanics and the first law of thermodynamics. [Assume that when a car is stopped, its kinetic energy \(\left(\frac{1}{2} m u^{2}\right)\) is totally converted to heat.]

From the following data, $$\begin{array}{r}\mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\\2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\\\\Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$calculate the enthalpy change for the reaction$$2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)$$

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?

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