Chapter 9, Problem 9.1P

The standard heat of the reaction

   $4N{H}_3\left(g\right)+O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O$

is $\Delta H_r^{°}=-904.7kJ$.

1. Briefly explain what that means. Your explanation may take the form "When

(specifyquantities of reactant species and their physical states) react to form

(quantities of product

species and their physical state), the change in enthalpy is

Is the reaction exothermic or endothermic at 25°C? Would you have to heat or cool the reactor to keep the temperature constant? What would the temperature do if the reactor ran adiabatically? What can you infer about the energy required to break the molecular bonds of the reactants and that released when the product bonds form?

What is $\Delta H_r^{°}$for

   $2N{H}_3\left(g\right)+\frac{5}{2}{O}_2\to 2NO\left(g\right)+3H_{2}O$

What is $\Delta H_r^{°}$

1. for

   $NO\left(g\right)+\frac{3}{2}{H}_{2}O\to N{H}_3\left(g\right)+\frac{5}{2}{O}_2$

1. Estimate the enthalpy change associated with the consumption of 340 g NH₃/s if the reactants and products are all at 25°C. (See Example 9.1-1.) What have you assumed about the reactor pressure? (You don't have to assume that it equals 1 atm.)
The values of $\Delta H_r^{°}$given in this problem apply to water vapor at 25°C and 1 atm, and yet the normal boiling point of water is 100°C. Can water exist as a vapor at 25°C and a total pressure of 1 atm? Explain your answer.

Interpretation Introduction

(a)

To explain:

The given equation $4N{H}_3(g)+5O_2\to 4NO(g)+6H_{2}O(g)$; $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$ in words.

Concept introduction:

When chemical reaction transforms the matter, there will be change in enthalpy in a system. This will happen when the products and reactants which involves in the chemical reaction are in their states. This is called standard heat or enthalpy of reaction.

Answer to Problem 9.1P

When 4 moles of gaseous ammonia and 5 moles of gaseous oxygen react to form 4 moles of gaseous nitric oxide and 6 moles of gaseous water, the change in enthalpy is $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$

Explanation of Solution

The standard heat of the reaction is,

$4N{H}_3(g)+5O_2\to 4NO(g)+6H_{2}O(g)$.

When 4 moles of gaseous ammonia and 5 moles of gaseous oxygen react to form 4 moles of gaseous nitric oxide and 6 moles of gaseous water, the change in enthalpy is $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$

Interpretation Introduction

(b)

To say:

Whether the reaction is exothermic or endothermic. To maintain the temperature constant, what would we do? If we ran the reactor adiabatically, what will be the state of the temperature?

Concept introduction:

A system will either absorbs heat or release heat to the surroundings. Exothermic reaction is the one where the heat gets released and endothermic reaction is the one where the heat gets absorbed. In exothermic, value of $\Delta H$ will be negative whereas in endothermic, value of $\Delta H$ will be positive.

Answer to Problem 9.1P

The reactor should be kept cool, to keep the temperature constant. The given reaction is exothermic. If the reactor is ran adiabatically, the temperature will get raise. The energy needed is low to break the reactant’s molecular bonds.

Explanation of Solution

We have $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$. As the change in enthalpy is negative, the reaction will be an exothermic one at 25ºC. To keep the temperature constant, the reactor should be kept cool. If the reactor is ran adiabatically i.e. there will be no heat exchange, the temperature would get raise. The energy required to break the molecular bonds of the reactants will be low than the energy released to form the product bonds. This is because of the reaction which is exothermic.

Interpretation Introduction

(c)

To find:

The $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r$ for the reaction $2N{H}_3(g)+\frac{5}{2}{O}_2\to 2NO(g)+3H_{2}O(g)$.

Concept introduction:

When chemical reaction transforms the matter, there will be change in enthalpy in a system. This will happen when the products and reactants which involves in the chemical reaction are in their states. This is called standard heat or enthalpy of reaction.

Answer to Problem 9.1P

Thus, the $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r\text{ is}-452.35kJ/mol$

Explanation of Solution

When we compare the reaction $2N{H}_3(g)+\frac{5}{2}{O}_2\to 2NO(G)+3H_{2}O(g)$ with $4N{H}_3(g)+5O_2\to 4NO(g)+6H_{2}O(g)$, it is clear that the reactants and products of the two reactions are same except the number of moles which get reduced into half.

So, for the reaction $4N{H}_3(g)+5O_2\to 4NO(g)+6H_{2}O(g)$, the change in enthalpy is $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$.

Therefore, for the reaction $2N{H}_3(g)+\frac{5}{2}{O}_2\to 2NO(G)+3H_{2}O(g)$, the change in enthalpy will be calculated by reducing the $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r$ by 2.

$\begin{array}{l}\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=\frac{-904.7}{2}kJ/mol\\ \Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-452.35kJ/mol\end{array}$

Interpretation Introduction

(d)

To find:

The $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r$ for the reaction $NO(g)+\frac{3}{2}{H}_{2}O(g)\to 2N{H}_3(g)+\frac{5}{4}{O}_2(g)$.

Concept introduction:

When chemical reaction transforms the matter, there will be change in enthalpy in a system. This will happen when the products and reactants which involves in the chemical reaction are in their states. This is called standard heat or enthalpy of reaction.

Answer to Problem 9.1P

Thus, the $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r\text{ is 226}\text{.2}kJ/mol$

Explanation of Solution

When we compare the reaction $NO(g)+\frac{3}{2}{H}_{2}O(g)\to 2N{H}_3(g)+\frac{5}{4}{O}_2(g)$ with $4N{H}_3(g)+5O_2\to 4NO(G)+6H_{2}O(g)$, it is clear that the reactants and products of the two reactions place get reversed and the number of moles gets reduced.

So, for the reaction $4N{H}_3(g)+5O_2\to 4NO(G)+6H_{2}O(g)$, the change in enthalpy is $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$.

Therefore, for the reaction $NO(g)+\frac{3}{2}{H}_{2}O(g)\to 2N{H}_3(g)+\frac{5}{4}{O}_2(g)$, the change in enthalpy will be calculated by reducing the $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r$ by 4.

$\begin{array}{l}\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=\frac{-904.7}{4}kJ/mol\\ \Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-226.2kJ/mol\end{array}$.

The reactants and products of the two reactions get reversed. So, the sign will also change.

$\therefore \Delta {\stackrel{\Lambda }{H^{\circ }}}_r=+226.2kJ/mol$

Interpretation Introduction

(e)

To estimate:

The enthalpy change and the reactor pressure.

Concept introduction:

When chemical reaction transforms the matter, there will be change in enthalpy in a system. This will happen when the products and reactants which involves in the chemical reaction are in their states. This is called standard heat or enthalpy of reaction.

Answer to Problem 9.1P

Thus, the change in enthalpy at ${25}^{\circ }C$ is, $\Delta \stackrel{\Lambda }{H}=4523kJ/s$

Explanation of Solution

For the reaction $4N{H}_3(g)+5O_2\to 4NO(G)+6H_{2}O(g)$, the change in enthalpy is $\Delta {\stackrel{\Lambda }{H^{\circ }}}_r=-904.7kJ/mol$.

For 4g of $N{H}_3$, the change in enthalpy is, $-904.7kJ/mol$.

Converting 1g of $N{H}_3$ will be $17g{\text{ of NH}}_3/mol$.

Converting 340g of $N{H}_3$ will be

$\frac{340N{H}_3/s}{17gofN{H}_3/mol}=20mol/s$.

The change n enthalpy is,

$\begin{array}{l}\Delta \stackrel{\Lambda }{H}=\frac{\frac{-904.7kJ/mol}{4molN{H}_3}}{20mol/s}=\frac{-904.7kJ/mol}{0.2}=4523.5\\ \Delta \stackrel{\Lambda }{H}=4523kJ/s\end{array}$.

The reactor pressure is same for reactants and products.

Interpretation Introduction

(f)

To explain:

Whether water exists as a vapor at ${25}^{\circ }C$ and a pressure of 1 atm.

Concept introduction:

When chemical reaction transforms the matter, there will be change in enthalpy in a system. This will happen when the products and reactants which involves in the chemical reaction are in their states. This is called standard heat or enthalpy of reaction.

Answer to Problem 9.1P

Thus, the water exists as a vapor at ${25}^{\circ }C$ and a pressure of 1 atm.

Explanation of Solution

Pure water will vapor at lower temperature. Therefore, water exists as a vapor at ${25}^{\circ }C$ and a pressure of 1 atm.