Question

Consider these reactions, where M represents a generic metal. 2M(s)+6HCl(aq)⟶2MCl3(aq)+3H2(g)ΔH1=−556.0 kJ HCl(g)⟶HCl(aq) ΔH2=−74.8 kJ H2(g)+Cl2(g)⟶2HCl(g) ΔH3=−1845.0 kJ MCl3(s)⟶MCl3(aq) ΔH4=−342.0 kJ Use the given information to determine the enthalpy of the reaction 2M(s)+3Cl2(g)⟶2MCl3(s)

Answer 1

The enthalpy change of a given reaction is $\boxed{-{\text{6111}}{\text{.8kJ}}}$.

Further Explanation:

This problem is based upon Hess’s Law. Hess’s law is utilized for calculating the enthalpy change of a reaction that can be obtained simply by summation of two or more reactions. In accordance with the Hess’s law, $\Delta H$of an overall reaction is obtained by adding the enthalpy change for each individual step reaction involved to obtain the overall reaction.

$\boxed{{{\Delta}}{H_{{\text{overall rxn}}}={\Delta}}{H_{\text{1}}+{{\Delta }}{H_{\text{2}}}+.......+{\Delta }}{H_{\text{n}}}}$

Enthalpy is defined as state function and therefore its value depends upon the initial and final state of system but not upon the path. This is the reason that the overall reaction can be simply obtained by adding or subtracting the enthalpy change of the individual steps utilized to get the final reaction.

Step 1: The enthalpy change of the following reaction is $\Delta {H_1}$.

$2{\text{M}}\left(s\right)+6{\text{HCl}}\left({aq}\right)\to2{\text{MC}}{{\text{l}}_3}\left({aq}\right)+3{{\text{H}}_2}\left(g\right)$              …… (1)

The value of $\Delta {H_1}$is$- {\text{556}}{\text{.0 kJ}}$.

Step 2: The enthalpy change of the following reaction is $\Delta {H_2}$.

${\text{HCl}}\left(g\right)\to{\text{HCl}}\left({aq}\right)$                                …… (2)

The value of $\Delta {H_{{\text{2}}}}$is $- {\text{74}}{\text{.8 kJ}}$.

Step 3: The enthalpy change of the following reaction is $\Delta {H_3}$.

${{\text{H}}_2}\left(g\right)+{\text{C}}{{\text{l}}_2}\left(g\right)\to2{\text{HCl}}\left(g\right)$                    …… (3)

The value of $\Delta {H_3}$is $- {\text{1845}}{\text{.0}}\;{\text{kJ}}$

Step 4: The enthalpy change of the following reaction is$\Delta {H_4}$.

${\text{MC}}{{\text{l}}_3}\left(s\right)\to{\text{MC}}{{\text{l}}_3}\left({aq}\right)$                            …… (4)

The value of $\Delta {H_4}$is $- {\text{342}}{\text{.0}}\;{\text{kJ}}$.

Step 5: Multiply the equation (2) by 6.

${\text{6HCl}}\left(g\right)\to6{\text{HCl}}\left({aq}\right)$                       …… (5)

Step 6: The enthalpy change for the reaction (5) is calculated as follows:

$\begin{aligned}\Delta{H_5}&=6\left({-{\text{74}}{\text{.8 kJ}}}\right)\\&=-{\text{448}}{\text{.8 kJ}}\\\end{aligned}$

Step 7: Multiply equation (3) by 3.

${\text{3}}{{\text{H}}_2}\left(g\right)+3{\text{C}}{{\text{l}}_2}\left(g\right)\to6{\text{HCl}}\left(g\right)$                 …… (6)

Step 8: The enthalpy change for the reaction (6) is calculated as follows:

$\begin{aligned}{\Delta }}{H_{\text{6}}}&={\text{3}}\left({-{\text{1845}}{\text{.0 kJ}}}\right)\\&=-{\text{5535}}{\text{.0 kJ}}\\\end{aligned}$

Step 9: Reverse and multiply equation (4) by 2.

$2{\text{MC}}{{\text{l}}_3}\left({aq}\right)\to{\text{2MC}}{{\text{l}}_3}\left(s\right)$                     …… (7)

The enthalpy change for the reaction (7) is calculated as follows:

$\begin{aligned}\Delta{H_7}&=-{\text{2}}\left({-{\text{342}}{\text{.0kJ}}}\right)\\&=+{\text{684}}{\text{.0 kJ}}\\\end{aligned}$

Add equation (1), (5),(6) and (7) to get the final equation.

$2{\text{M}}\left(s\right)+3{\text{C}}{{\text{l}}_2}\left(g\right)\to2{\text{MC}}{{\text{l}}_3}\left(s\right)$                         ……(8).

The enthalpy change of the following reaction is $\Delta {H_8}$

Step 9: The expression to calculate$\Delta {H_8}$is as follows:

$\Delta {H_8}=\Delta {H_1}+\Delta {H_5}+\Delta {H_6}+\Delta{{\text{H}}_7}$            …… (9)

Substitute $- {\text{556}}{\text{.0 kJ}}$for $\Delta {H_1}$, $- {\text{448}}{\text{.8 kJ}}$for $\Delta {H_5}$, $- {\text{5535}}{\text{.0 kJ}}$ for $\Delta {H_6}$and $+{\text{684}}{\text{.0 kJ}}$for $\Delta{{\text{H}}_7}$in the equation (9).

$\begin{aligned}\Delta{H_8}&=\left({-{\text{556}}{\text{.0 kJ}}}\right)+\left({-448.8\;{\text{kJ}}}\right)+\left({-5535.0\;{\text{kJ}}}\right)+\left({+{\text{684}}{\text{.0 kJ}}}\right)\\&=-{\text{5849}}{\text{.8}}\;{\text{kJ}}\\\end{aligned}$

Hence, the enthalpy of the given reaction (8) is.${\mathbf{-5849}}{\mathbf{.8}}\;{\mathbf{kJ}}$.

Learn more:

1. Oxidation and reduction reaction brainly.com/question/2973661

2. Calculation of moles of HClhttps://brainly.com/question/5950133

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Thermodynamics

Keywords: Hess Law, enthalpy, overall reaction, adding, state function, initial state, final state, HCl, M, ${\text{C}}{{\text{l}}_2}$,${\text{MC}}{{\text{l}}_3}$,-5849.8 kj, -556.0 kj and -448.8 kj.

Answer 2

Chemical reaction 1: 2M(s)+6HCl(aq)⟶ 2MCl3(aq)+3H2(g); ΔH1=−556.0 kJ. Chemical reaction 2: HCl(g) ⟶ HCl(aq); ΔH2=−74.8 kJ.
Chemical reaction 3: H2(g)+Cl2(g) ⟶ 2HCl(g); ΔH3=−1845.0 kJ
Chemical reaction 4: MCl3(s) ⟶ MCl3(aq); ΔH4=−342.0 kJ.
Chemical reaction 5: 2M(s)+3Cl2(g) ⟶ 2MCl3(s); ΔH5 = ?.ΔH5 = ΔH1 + 6·ΔH2 + 3·ΔH3 - 2·ΔH4.
ΔH5 = -550 kJ + 6·(-74,8 kJ) + 3·(-1845 kJ) - 2·(-342 kJ).
ΔH5 = -550 kJ - 448,8 kJ - 5535 kJ + 684 kJ.
ΔH5 = -5849,8 kJ.