Chapter 5, Problem 108SCQ

(a)

Interpretation Introduction

Interpretation:

To show the net result is the decomposition of water

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity, $\Delta T$= change in temperature.

The standard molar enthalpy of formation is the enthalpy change ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ is the enthalpy change for the formation of $\text{1mol}$ of a compound directly from its component elements in their standard states. And is given by

                                 ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

To show the net result is the decomposition of water.

Equation $\text{1-2}\text{=A}$

$\text{A=}$${\text{CaBr}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+Hg}\to \text{CaO}\text{+}{\text{HgBr}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

Equation $\text{A-3=B}$

${\text{H}}_{\text{2}}\text{O}\text{+}\text{Hg}\to \text{HgO}\text{+}{\text{H}}_{\text{2}}$

Equation $\text{4-B}$ is the required equation.

${\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}\text{+}{\text{1/2O}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

If $\text{1000Kg}$ of water is used then identify the mass of ${\text{H}}_{\text{2}}$ produced

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

                                                  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity,$\Delta T$= change in temperature.

The standard molar enthalpy of formation is the enthalpy change ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ is the enthalpy change for the formation of $\text{1mol}$ of a compound directly from its component elements in their standard states. And is given by

                                 ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

The mass of ${\text{H}}_{\text{2}}$ produced

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{0-(-285}\text{.8kJ/mol)}$ =$\text{285}\text{.8kJ/mol}$

Substitute the values in, $\text{q=C×m×ΔT}$, as

$\text{q}\text{=}$$\text{-285}\text{.8kJ/mol}\text{×}\text{1000Kg}$= $\text{28kJ}$

$\text{28kJ×}\frac{\text{1mol}{\text{H}}_{\text{2}}}{\text{285}\text{.8kJ}}\text{×}\frac{\text{1000Kg}}{\text{1mol}}$ =$\text{97}\text{.97Kg}$

The reaction is ${\text{H}}_{\text{2}}\text{O}\to \text{1/}{\text{2O}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

(c)

Interpretation Introduction

Interpretation:

The enthalpy of reaction has to be calculated.

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

                                                  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity,$\Delta T$= change in temperature.

The standard molar enthalpy of formation is the enthalpy change ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ is the enthalpy change for the formation of $\text{1mol}$ of a compound directly from its component elements in their standard states. And is given by

                                 ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

Given reaction is:

  ${\text{CaBr}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\text{O}\text{+Hg}\to \text{CaO}\text{+}{\text{HgBr}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}$

 Substitute the values in ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$ as,

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$= $\text{(-635}\text{.5}\text{+}\text{-169}\text{.5 kJ/mol}\text{+}\text{0)}$ - $\text{( -675 kJ/mol}\text{+}\text{-285}\text{.8kJ/mol)}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{155}\text{.8kJ/mol}$

${\text{H}}_{\text{2}}\text{O}\text{+}\text{Hg}\to \text{HgO}\text{+}{\text{H}}_{\text{2}}$

Substitute the values in ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$ as,

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{(-90}\text{.7kJ/mol)}\text{-}\text{(-285}\text{.8kJ/mol)}$ =$\text{195}\text{.1kJ/mol}$

${\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}\text{+}{\text{1/2O}}_{\text{2}}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{285}\text{.8kJ/mol}$

(d)

Interpretation Introduction

Interpretation:

Comment on the feasibility of using such a series of reaction to produce ${\text{H}}_{\text{2}}$

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1K.Energy gained or lost can be calculated using the below equation.

  $\text{q=C×m×ΔT}$

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity,$\Delta T$= change in temperature.

The standard molar enthalpy of formation is the enthalpy change ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$ is the enthalpy change for the formation of $\text{1mol}$ a compound directly from its component elements in their standard states. And is given by

                                 ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

Since the enthalpy is positive the reaction is not feasible