Chapter 5, Problem 83GQ

(a)

Interpretation Introduction

Interpretation:

The enthalpy change for the reaction using standard enthalpies has to be determined

Concept Introduction:

The standard enthalpy of formation is the enthalpy change for the formation of 1mol of the compound directly from its component elements in their standard states.

Enthalpy change for the reaction ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

Given,

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(C)}\text{=}\text{ 0 KJ/mol }$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{(H}}_{\text{2}}\text{O)}$=$\text{-241}\text{.8kJ/mol}$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(CO)}$=$\text{-110}\text{.54kJ/mol}$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{(H}}_{\text{2}}\text{)}$=$\text{0J/mol}$

Enthalpy change for the reaction ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

  ${\text{Δ}}_{\text{r}}\text{H° =}\text{(-110}\text{.54}\text{+}\text{0)-}\text{(0}\text{+}\text{-241}\text{.8)}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{131}\text{.26 kJ/mol}$

So, the change in enthalpy of the reaction is $\text{131}\text{.3}\text{kJ/mol}$

(b)

Interpretation Introduction

Interpretation:

The nature of the reaction has to be identified.

Concept Introduction:

The standard enthalpy of formation is the enthalpy change for the formation of 1mol of the compound directly from its component elements in their standard states.

Enthalpy change for the reaction ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

Given,

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(C)}\text{=}\text{ 0 KJ/mol }$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{(H}}_{\text{2}}\text{O)}$=$\text{-241}\text{.8kJ/mol}$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(CO)}$=$\text{-110}\text{.54kJ/mol}$

${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}{\text{(H}}_{\text{2}}\text{)}$=$\text{0J/mol}$

Enthalpy change for the reaction ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

  ${\text{Δ}}_{\text{r}}\text{H° =}\text{(-110}\text{.54}\text{+}\text{0)-}\text{(0}\text{+}\text{-241}\text{.8)}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{131}\text{.26 kJ/mol}$

The change in enthalpy is $\text{+131}\text{.3}\text{kJ/mol}$, so it is endothermic reaction.

(c)

Interpretation Introduction

Interpretation:

The enthalpy change if $\text{1000}\text{.0}\text{kg}$ of carbon converted to water gas has to be calculated.

Concept Introduction:

The standard enthalpy of formation is the enthalpy change for the formation of 1mol of the compound directly from its component elements in their standard states.

Enthalpy change for the reaction ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

Explanation of Solution

The change in enthalpy is $\text{+131}\text{.3}\text{kJ/mol}$,

Heat evolved when 1000Kg of carbon is converted to coal:

$\text{1000Kg}\text{×}\frac{\text{1000g}}{\text{1kg}}\text{×}\frac{\text{1mol}}{\text{12}\text{.01g}}\text{×}\frac{\text{131}\text{.31kJ}}{\text{1mol}}$=$\text{1}\text{.0932}\text{×}{\text{10}}^{\text{7}}\text{kJ}$

So, the enthalpy change if $\text{1000}\text{.0}\text{kg}$ of carbon converted to water gas is $\text{1}\text{.0932}\text{×}{\text{10}}^{\text{7}}\text{kJ}$