Question

Use the values of DH8f in Appendix 4 to calculate DH8 for the following reactions.

a.

b.${\mathbf{Ca}}_{\mathbf{3}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\left(l\right)\mathbf{\to }\mathbf{3}{\mathbf{CaSO}}_{\mathbf{4}}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\left(l\right)$

c.${\mathbf{NH}}_{\mathbf{3}}\left(g\right)\mathbf{+}\mathbf{HCl}\left(g\right)\mathbf{\to }{\mathbf{NHCl}}_{\mathbf{4}}\left(s\right)$

d.

e.${\mathbf{SiCl}}_{\mathbf{4}}\left(l\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }{\mathbf{SiO}}_{\mathbf{2}}\left(s\right)\mathbf{+}\mathbf{4}\mathbf{HCl}\left(\mathrm{aq}\right)$

f.$\mathbf{MgO}\left(s\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)\mathbf{\to }\mathbf{Mg}{\left(\mathrm{OH}\right)}_{\mathbf{2}}\left(s\right)$

Short answer

a.$∆{\mathrm{H}}^{°}=-940\mathrm{kJ}$

b.$∆{\mathrm{H}}^{°}=-265\mathrm{kJ}$

c.$∆{\mathrm{H}}^{°}=-176\mathrm{kJ}$

d.$∆{\mathrm{H}}^{°}=-1235\mathrm{kJ}$

e.$∆{\mathrm{H}}^{°}=-320\mathrm{kJ}$

f.$∆{\mathrm{H}}^{°}=-37\mathrm{kJ}$

Step-by-step solution

Calculate the enthalpy change for the given reaction.

The balanced chemical equation for this illustrated reaction is written below:

$2{\mathrm{NH}}_3\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{CH}}_4\left(\mathrm{g}\right)\to 2\mathrm{HCN}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The value of$∆\mathrm{H}$ for the above reactioncan be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left(\mathrm{HCN}\left(\mathrm{g}\right)\right)+6\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)\right]-\\ & & \left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{NH}}_3\left(\mathrm{g}\right)\right)+3\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)+2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{CH}}_4\left(\mathrm{g}\right)\right)\right]\\ & =& \left[2\mathrm{mol}\times 135.1\mathrm{kJ}/\mathrm{mol}+6\mathrm{mol}\times -242\mathrm{kJ}/\mathrm{mol}\right]-\\ & & \left[2\mathrm{mol}\times -46\mathrm{kJ}/\mathrm{mol}+3\mathrm{mol}\times 0+2\mathrm{mol}\times -75\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -940\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for $2{\mathrm{NH}}_3\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)+2{\mathrm{CH}}_4\left(\mathrm{g}\right)\to 2\mathrm{HCN}\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$is $∆{\mathrm{H}}^{°}=-940\mathrm{kJ}$.

Calculate the enthalpy change for the given reaction.

The balanced chemical equation for this reaction is written below:

${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)+3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to 3{\mathrm{CaSO}}_4\left(\mathrm{s}\right)+2{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)$

The value of for the above reactioncan be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[3\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{CaSO}}_4\left(\mathrm{s}\right)\right)+2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)\right)\right]\\ & & -\left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)\right)+3\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\right)\right]\\ & =& \left[3\mathrm{mol}\times -1433\mathrm{kJ}/\mathrm{mol}+2\mathrm{mol}\times -1267\mathrm{kJ}/\mathrm{mol}\right]\\ & & -\left[1\mathrm{mol}\times -4126\mathrm{kJ}/\mathrm{mol}+3\mathrm{mol}\times -814\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -265\mathrm{kJ}\end{array}$

Hence, the $∆\mathrm{H}$for ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\left(\mathrm{s}\right)+3{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to 3{\mathrm{CaSO}}_4\left(\mathrm{s}\right)+2{\mathrm{H}}_3{\mathrm{PO}}_4\left(\mathrm{l}\right)$is $∆{\mathrm{H}}^{°}=-265\mathrm{kJ}$.

Calculate the enthalpy change for the given reaction.

The balanced chemical equation for this reaction is written below:

${\mathrm{NH}}_3\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{g}\right)\to {\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)$

The value of for the above reaction can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)\right)\right]-\left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{NH}}_3\left(\mathrm{g}\right)\right)+1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left(\mathrm{HCl}\left(\mathrm{g}\right)\right)\right]\\ & =& \left[1\mathrm{mol}\times -314\mathrm{kJ}/\mathrm{mol}\right]-\left[1\mathrm{mol}\times -46\mathrm{kJ}/\mathrm{mol}+1\mathrm{mol}\times -92\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -176\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for ${\mathrm{NH}}_3\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{g}\right)\to {\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)$is $∆{\mathrm{H}}^{°}=-176\mathrm{kJ}$.

 Step 4: Calculate the enthalpy change for the given reaction.

The balanced chemical equation for this illustrated reaction is written below:

${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The value of for the above reaction can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{CO}}_2\left(\mathrm{g}\right)\right)+3\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\right)\right]\\ & & -\left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)\right)+3\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{O}}_2\left(\mathrm{g}\right)\right)\right]\\ & =& \left[2\mathrm{mol}\times -393.5\mathrm{kJ}/\mathrm{mol}+2\mathrm{mol}\times -242\mathrm{kJ}/\mathrm{mol}\right]\\ & & -\left[1\mathrm{mol}\times -278\mathrm{kJ}/\mathrm{mol}+3\mathrm{mol}\times 0\right]\\ & =& -1235\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{l}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{CO}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$is $∆{\mathrm{H}}^{°}=-1235\mathrm{kJ}$.

Calculate the enthalpy change for the given reaction.

The balanced chemical equation for this reaction is written below:

${\mathrm{SiCl}}_4\left(\mathrm{l}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{SiO}}_2\left(\mathrm{s}\right)+4\mathrm{HCl}\left(\mathrm{aq}\right)$

The value of for the above reaction can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{SiO}}_2\left(\mathrm{s}\right)\right)+4\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left(\mathrm{HCl}\left(\mathrm{aq}\right)\right)\right]\\ & & -\left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{SiCl}}_4\left(\mathrm{l}\right)\right)+2\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)\right]\\ & =& \left[1\mathrm{mol}\times -911\mathrm{kJ}/\mathrm{mol}+4\mathrm{mol}\times -67\mathrm{kJ}/\mathrm{mol}\right]\\ & & -\left[1\mathrm{mol}\times -687\mathrm{kJ}/\mathrm{mol}+2\mathrm{mol}\times -286\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -320\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for ${\mathrm{SiCl}}_4\left(\mathrm{l}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{SiO}}_2\left(\mathrm{s}\right)+4\mathrm{HCl}\left(\mathrm{aq}\right)$is $-320\mathrm{kJ}$.

Calculate the enthalpy change for the given reaction.

The balanced chemical equation for this reaction is written below:

$\mathrm{MgO}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)$

The value of for the above reaction can be calculated as follows:

$\begin{array}{rcl}∆\mathrm{H}& =& \left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left(\mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)\right)\right]\\ & & -\left[1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left(\mathrm{MgO}\left(\mathrm{s}\right)\right)+1\mathrm{mol}\times ∆{\mathrm{H}}_{\mathrm{f}}^{°}\left({\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\right)\right]\\ & =& \left[3\mathrm{mol}\times -925\mathrm{kJ}/\mathrm{mol}\right]-\left[1\mathrm{mol}\times -602\mathrm{kJ}/\mathrm{mol}+3\mathrm{mol}\times -286\mathrm{kJ}/\mathrm{mol}\right]\\ & =& -37\mathrm{kJ}\end{array}$

Hence, the$∆\mathrm{H}$ for $\mathrm{MgO}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \mathrm{Mg}{\left(\mathrm{OH}\right)}_2\left(\mathrm{s}\right)$is $∆{\mathrm{H}}^{°}=-37\mathrm{kJ}$.