Chemistry, 19.09.2019 18:30 woodfordmaliky

Consider these reactions where m represents a generic metal
1. 2m(s) +6hcl(aq) --> 2mcl3(aq)+3h2(g) (deltah)= -725.0 kj
2. hcl(> hcl(aq) (deltah)= -74.8kj 3. h2(g)+cl2(g) --> 2hcl(g) (deltah)=-1845.0kj 4. mcl3(s) --> mcl3(aq) (deltah)= -476.0kj use the information above to determine the enthalpy of the following reaction.
2m(s)+3cl2(g) > 2mcl3(s) (deltah) = kj

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Answer from: anonymousanon

The enthalpy of the given reaction is $\boxed{ - 4383.2{\text{ kJ}}}$

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.

$\Delta\text{H}_{\text{overall rxn}}=\Delta\text{H}_{1}+\Delta\text{H}_{2}+......+\Delta\text{H}_{n}$

Enthalpy is defined as a 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 .

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

The value of $\Delta{H_1}$ is -725 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 $\Delta{H_2}$ is -74.5 kJ

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

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

The value of $\Delta {H_3}$ is -1845 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 -476 kJ.

Step 5: The enthalpy change of the following reaction is $\Delta {H_5}$ .

${\text{2M}}\left(s\right)+3{\text{C}}{{\text{l}}_2}\left(g\right)\to2{\text{MC}}{{\text{l}}_3}\left(s\right)$ …… (5)

Step 6: Multiply equation (2) by 6, equation (3) by 3 and equation (4) by 3. Add the resulting equation (3) and equation (1). Subtract the resulting equation (2) and equation (4). The final equation obtained is as follows:

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

Step 7: The expression to calculate $\Delta {H_5}$ is as follows:

$\Delta{H_5}=\Delta {H_1}-6\left( {\Delta{H_2}}\right)+3\left({\Delta {H_3}}\right)- 3\left( {\Delta {H_4}} \right)$ ...... (7)

Substitute -725 kJ for $\Delta {H_1}$ , -74.8 kJ for $\Delta {H_2}$ and -1845 kJ for $\Delta {H_3}$ and -476 kJ for $\Delta {H_4}$ in equation (7).

$\begin{aligned}\Delta {H_5}&=-{\text{725 kJ}}-6\left({-{\text{74}}{\text{.8 kJ}}}\right)+3\left({ - {\text{1845 kJ}}} \right)-3\left({ - {\text{476 kJ}}} \right)\\&=-{\text{725 kJ}}+{\text{448}}{\text{.8 kJ}}-{\text{5535 kJ}}+{\text{1428 kJ}}\\&=-{\text{4383}}{\text{.2 kJ}}\\\end{aligned}$

Hence, the enthalpy of ${\mathbf{2M}}\left({\mathbf{s}}\right){\mathbf{ + 3C}}{{\mathbf{l}}_{\mathbf{2}}}\left( {\mathbf{g}}\right)\to{\mathbf{2MC}}{{\mathbf{l}}_{\mathbf{3}}}\left( {\mathbf{s}}\right)$ is -4383.2 kJ.