Chapter 5, Problem 5.90QE

(a)

Interpretation Introduction

Interpretation:

Number of moles of $CO$ and $H_2$ in $10g$ water gas has to be determined.

Concept introduction:

Number of moles can be determined using the given equation,

  $\text{n}\text{= }\frac{\text{m}}{\text{M}}$

Here, n is the number of moles

  M is the Molar mass

    m is the Mass

Answer to Problem 5.90QE

$10g$ water gas has $\text{0}\text{.333 mol CO}$ and $\text{0}\text{.333 mol }{H}_2$.

Explanation of Solution

Mass of one mole of water gas is $15.01g$.

Number of moles of water gas is determined as follows,

  $\begin{array}{c}n\text{}\text{= }\frac{Mass}{Molar\text{}mass}\\ \\ =\frac{10.0\text{ g}}{15.01\text{ g/mol}}\\ \\ =\text{ 0}\text{.666 mol}\end{array}$

Ratio of $CO$ and $H_2$ in water gas is $1:\text{ 1}$. It implies that equal amount of $CO$ and $H_2$ is present in water gas. Therefore, 1 mole of water gas has $0.50\text{ mol }CO$ and $0.50\text{ mol }{H}_2$.

Number of moles of $CO$ and $H_2$ in $10g$ water gas it determine as follows,

  $\begin{array}{c}{n}_{CO}\text{}\text{= 0}\text{.666 mol water gas}\times \frac{0.50\text{ mol CO}}{\text{1 mol water gas}}\\ \\ =\text{ 0}\text{.333 mol CO}\\ \\ n_{H_2}\text{}\text{= 0}\text{.666 mol water gas}\times \frac{0.50\text{ mol }{H}_2}{\text{1 mol water gas}}\\ \\ =\text{ 0}\text{.333 mol }{H}_2\end{array}$

$10g$ water gas has $\text{0}\text{.333 mol CO}$ and $\text{0}\text{.333 mol }{H}_2$.

(b)

Interpretation Introduction

Interpretation:

Enthalpy change when $\text{10}\text{.0 g}$ water gas is burned should be calculated.

Concept introduction:

According to Hess’s Laws, change in enthalpy in an overall reaction can be calculated from the change in enthalpy of other reactions.

• If an equation is obtained by the addition of more than one thermochemical equations, then the enthalpy change of that equation is the sum of change in enthalpy of all the equations added.
• If an equation is the reverse direction of a thermochemical equation, then the change in enthalpy has same numerical value but opposite sign.
• The enthalpy change depends on the mass of reacting substance. If the coefficients present in an equation is multiplied with a factor, then change in enthalpy also should be multiplied with that same factor.

Answer to Problem 5.90QE

Enthalpy change when $\text{10}\text{.0 g}$ water gas is burned is $-\text{379 kJ}$.

Explanation of Solution

Mass of one mole of water gas is $15.01g$.

Number of moles of water gas is determined as follows,

  $\begin{array}{c}n\text{}\text{= }\frac{Mass}{Molar\text{}mass}\\ \\ =\frac{10.0\text{ g}}{15.01\text{ g/mol}}\\ \\ =\text{ 0}\text{.666 mol}\end{array}$

Given reactions are shown below,

  $\begin{array}{l}\text{2 CO}\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\to }\text{ 2}{\text{CO}}_{\text{2}}\left(g\right)\Delta H=-566kJ\left(1\right)\\ \\ {\text{2 H}}_2\left(g\right){\text{ + O}}_{\text{2}}\left(g\right)\underset{}{\to }\text{ 2}{\text{H}}_{\text{2}}O\left(l\right)\Delta H=-\text{571}\text{.7}kJ\left(2\right)\end{array}$

Balanced chemical equation for the burning of water gas in air is shown below,

  $CO\left(g\right){\text{+ H}}_2\left(g\right)+O_2\left(g\right)\underset{}{\to }C{O}_2\left(g\right){\text{ + H}}_{2}O\left(3\right)$

From the above equations, it is clear equation (3) can be obtained by adding equation (1) divided by 2 and equation (2) divided by 2.

  $\begin{array}{l}\text{}\text{ CO}\left(g\right)\text{ + }\frac{1}{2}{\text{O}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{CO}}_{\text{2}}\left(g\right)\\ \\ \underset{\_}{{\text{ H}}_2\left(g\right)\text{ + }\frac{1}{2}{\text{O}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{H}}_{\text{2}}O\left(l\right)}\\ \\ CO\left(g\right){\text{+ H}}_2\left(g\right)+O_2\left(g\right)\underset{}{\to }C{O}_2\left(g\right){\text{ + H}}_{2}O\end{array}$

According to Hess’s Law, when the direction of reaction reverses, then the enthalpy change will get opposite sign. Also, when the number of reactants and products are multiplied or divided by a factor, then the enthalpy change must also be multiplied or divided by the same factor.

$\Delta H$ for the combustion of 1 mole of water gas can be determined as follows,

  $\begin{array}{c}\Delta H_{eqn\text{ 3}}=\left(\frac{\Delta H_{eqn\text{ 1}}}{2}\right)+\left(\frac{\Delta H_{eqn\text{ 2}}}{2}\right)\\ \\ =\left(\frac{-\text{566 kJ}}{2}\right)+\left(\frac{-\text{571}\text{.7 kJ}}{2}\right)\\ \\ =-\text{568}\text{.9 kJ}\end{array}$

Enthalpy change for the combustion of 1 mole of water gas is $-568.9\text{ kJ}$. Hence enthalpy change when $\text{10}\text{.0 g}$ water gas is burned can be determined as follows,

  $\begin{array}{c}\Delta H=\text{ 0}\text{.666 mol}\times \frac{-568.9\text{ kJ}}{1\text{ mol}}\\ \\ =-\text{379 kJ}\end{array}$

Enthalpy change when $\text{10}\text{.0 g}$ water gas is burned is $-\text{379 kJ}$.