Question

Consider these reactions, where M represents a generic metal

1.) 2M(s)+6HCl(aq) --> 2MCl3(aq)+3H2(g) H=-748.0 kJ

2.) HCl(g) --> HCl(aq)  H=-74.8

3.) H2(g)+Cl2(g) --> 2HCl(g)  H=-1845

4.) MCl3(s) --> MCl3(aq)   H=-317.0

Using the information above to determine the enthalpy of the following reaction

2M(s)+3Cl2(g) --> 2MCl3(s)

H=______kJ

Answer 1

General guidance

Concepts and reason

To find the total enthalpy change of a reaction, perform the arithmetic operations on the given individual synthetic reactions to obtain it. It is then followed by performing the same set of operations on the corresponding $\\Delta H$ values.

Fundamentals

Hess\u2019s law: According to the Hess\u2019s law, also known as Hess\u2019s law of constant heat summation, the total enthalpy change for a particular reaction during the complete course of time is same irrespective of the path it follows.

The standard enthalpy change for a given chemical reaction can be calculated by using the following expression:

$\\Delta {\\rm{H}}_f^ \\circ \\left( {{\\rm{rxn}}} \\right)\\;{\\rm{ = }}\\;{\\rm{\\Delta H}}_f^ \\circ \\left( {{\\rm{products}}} \\right)\\; - \\;{\\rm{\\Delta H}}_f^ \\circ \\left( {{\\rm{reactants}}} \\right)$

Here, $\\Delta {\\rm{H}}_f^ \\circ$ represents the standard enthalpy change of formation.

Using the Hess\u2019s law, total enthalpy change of the reaction can be calculated by summing the enthalpy change of its individual synthetic steps.

Step-by-step

Step 1 of 3

The given reactions are as follows:

${\\rm{2M}}\\;\\left( s \\right)\\;{\\rm{ + }}\\;{\\rm{6HCl}}\\;\\left( {aq} \\right) \\to {\\rm{2MC}}{{\\rm{l}}_{\\rm{3}}}\\left( {aq} \\right){\\rm{ + 3}}{{\\rm{H}}_{\\rm{2}}}\\left( g \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\; - 748.0\\;{\\rm{kJ}}$ \u2026\u2026(1)

${\\rm{HCl}}\\;\\left( g \\right)\\; \\to {\\rm{HCl}}\\left( {aq} \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\; - 74.8\\;{\\rm{kJ}}$ \u2026\u2026(2)

${{\\rm{H}}_2}\\left( g \\right)\\;{\\rm{ + }}\\;{\\rm{C}}{{\\rm{l}}_2}\\;\\left( g \\right) \\to {\\rm{2HCl}}\\left( g \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\; - 1845\\;{\\rm{kJ}}$ \u2026\u2026(3)

${\\rm{MC}}{{\\rm{l}}_3}\\;\\left( s \\right) \\to {\\rm{MC}}{{\\rm{l}}_{\\rm{3}}}\\left( {aq} \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\; - 317.0\\;{\\rm{kJ}}$ \u2026\u2026(4)

Perform the following steps to form equation:

$2{\\rm{M}}\\;\\left( s \\right) + \\;3{\\rm{C}}{{\\rm{l}}_{\\rm{2}}}\\left( g \\right) \\to 2{\\rm{MC}}{{\\rm{l}}_{\\rm{3}}}\\left( s \\right)$

Keep equation (1) as it is without any change and multiply equation (3) by 3. Also, multiply equation (2) by 6 and equation (4) by 2 and then reversing it as follows:

${\\rm{2M}}\\;\\left( s \\right)\\;{\\rm{ + }}\\;{\\rm{6HCl}}\\;\\left( {aq} \\right) \\to {\\rm{2MC}}{{\\rm{l}}_{\\rm{3}}}\\left( {aq} \\right){\\rm{ + 3}}{{\\rm{H}}_{\\rm{2}}}\\left( g \\right)$ \u2026\u2026(5)

${\\rm{6HCl}}\\;\\left( g \\right)\\; \\to 6{\\rm{HCl}}\\left( {aq} \\right)$ \u2026\u2026(6)

${\\rm{3}}{{\\rm{H}}_2}\\left( g \\right)\\;{\\rm{ + }}\\;3{\\rm{C}}{{\\rm{l}}_2}\\;\\left( g \\right) \\to 6{\\rm{HCl}}\\left( g \\right)$ \u2026\u2026(7)

${\\rm{2MC}}{{\\rm{l}}_{\\rm{3}}}\\left( {aq} \\right) \\to 2{\\rm{MC}}{{\\rm{l}}_3}\\;\\left( s \\right)$ \u2026\u2026(8)

Explanation

To form equation $2{\\rm{M}}\\;\\left( s \\right) + \\;3{\\rm{C}}{{\\rm{l}}_{\\rm{2}}}\\left( g \\right) \\to 2{\\rm{MC}}{{\\rm{l}}_{\\rm{3}}}\\left( s \\right)$, equation (1) is kept as it is and equation (3) is multiplied by 3 to obtain the same reactants as that of former reaction.

The extra reactants and products are removed by multiplying equation (2) by 6 and reversing equation (4) and then multiplying it by 2.

Step 2 of 3

Perform the same operations with $\\Delta H$ values as in equations (5), (6), (7), and (8) respectively.

${\\rm{2M}}\\;\\left( s \\right)\\;{\\rm{ + }}\\;{\\rm{6HCl}}\\;\\left( {aq} \\right) \\to {\\rm{2MC}}{{\\rm{l}}_{\\rm{3}}}\\left( {aq} \\right){\\rm{ + 3}}{{\\rm{H}}_{\\rm{2}}}\\left( g \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\; - 748.0\\;{\\rm{kJ}}$ \u2026\u2026(9)

$\\begin{array}{l}\\\\{\\rm{6HCl}}\\;\\left( g \\right)\\; \\to 6{\\rm{HCl}}\\left( {aq} \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\;6 \\times \\left( { - 74.8\\;{\\rm{kJ}}} \\right)\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = - \\;448\\;{\\rm{kJ}}\\\\\\end{array}$ \u2026\u2026(10)

$\\begin{array}{l}\\\\{\\rm{3}}{{\\rm{H}}_2}\\left( g \\right)\\;{\\rm{ + }}\\;3{\\rm{C}}{{\\rm{l}}_2}\\;\\left( g \\right) \\to 6{\\rm{HCl}}\\left( g \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\;3 \\times \\left( { - 1845\\;{\\rm{kJ}}} \\right)\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\; - 5535\\;{\\rm{kJ}}\\\\\\end{array}$ \u2026\u2026(11)

$\\begin{array}{l}\\\\{\\rm{2MC}}{{\\rm{l}}_{\\rm{3}}}\\left( {aq} \\right) \\to \\;2{\\rm{MC}}{{\\rm{l}}_3}\\;\\left( s \\right)\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\Delta H\\; = \\;2 \\times \\left( { + 317.0\\;{\\rm{kJ}}} \\right)\\\\\\\\\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\; = \\;634\\;\\;{\\rm{kJ}}\\\\\\end{array}$ \u2026\u2026(12)

Explanation

Equation (9) is same as equation (5) in Step 1 as no arithmetic calculation was performed in that step.

Equation (10) is obtained by multiplying $\\Delta H$ of equation (6) by 6.

Equation (11) is obtained by multiplying $\\Delta H$ of equation (7) by 3.

Equation (12) is obtained by changing the sign of $\\Delta H$ and multiplying it by 2. The sign is changed as equation (4) is reversed.

Step 3 of 3

Calculate the enthalpy of reaction for the final reaction as follows:

$\\begin{array}{c}\\\\{\\rm{Enthalpy}}\\;{\\rm{of}}\\;{\\rm{Reaction}}\\;{\\rm{ = }}\\; - 748.0\\;{\\rm{kJ}}\\; + \\;\\left( { - \\;448\\;{\\rm{kJ}}} \\right) + \\left( { - \\;5535\\;{\\rm{kJ}}} \\right) + \\left( { + \\;634\\;{\\rm{kJ}}} \\right)\\\\\\\\ = \\; - 6097.8\\;{\\rm{kJ}}\\\\\\end{array}$

The enthalpy of reaction, $2{\\rm{M}}\\;\\left( s \\right) + \\;3{\\rm{C}}{{\\rm{l}}_{\\rm{2}}}\\left( g \\right) \\to 2{\\rm{MC}}{{\\rm{l}}_{\\rm{3}}}\\left( s \\right)$, is .

Explanation

Total enthalpy of the reaction is calculated by adding $\\Delta H$ of individual reaction which are obtained in step 3.

The negative sign of total enthalpy of the reaction indicates that the product formation is favorable.

Answer

The enthalpy of reaction, $2{\\rm{M}}\\;\\left( s \\right) + \\;3{\\rm{C}}{{\\rm{l}}_{\\rm{2}}}\\left( g \\right) \\to 2{\\rm{MC}}{{\\rm{l}}_{\\rm{3}}}\\left( s \\right)$, is .

Answer only

The enthalpy of reaction, $2{\\rm{M}}\\;\\left( s \\right) + \\;3{\\rm{C}}{{\\rm{l}}_{\\rm{2}}}\\left( g \\right) \\to 2{\\rm{MC}}{{\\rm{l}}_{\\rm{3}}}\\left( s \\right)$, is .