Question

calculate the enthalpy for this reaction.
2hg + cl₂ → hg₂cl₂
use the following thermochemical equations.

1. hgcl₂ → hg + cl₂ δh₁ = +224 kj
2. hg + hgcl₂ → hg₂cl₂ δh₂ = −41 kj

δhᵣₓₙ = ? kj
enter either a + or - sign and the magnitude.
use significant figures.

Explanation:

Step1: Reverse the first equation

To get \( Hg + Cl_2
ightarrow HgCl_2 \), we reverse the first equation \( HgCl_2
ightarrow Hg + Cl_2 \) with \( \Delta H_1 = +224 \, \text{kJ} \). Reversing the reaction changes the sign of \( \Delta H \), so the new \( \Delta H \) for the reversed reaction is \( -224 \, \text{kJ} \).

Step2: Add the reversed first equation and the second equation

Now we have two equations:

1. \( Hg + Cl_2

ightarrow HgCl_2 \) with \( \Delta H = -224 \, \text{kJ} \)

1. \( Hg + HgCl_2

ightarrow Hg_2Cl_2 \) with \( \Delta H_2 = -41 \, \text{kJ} \)

Adding these two equations together: \( (Hg + Cl_2) + (Hg + HgCl_2)
ightarrow HgCl_2 + Hg_2Cl_2 \). Simplifying the reactants and products, the \( HgCl_2 \) on the left and right cancels out, and we get \( 2Hg + Cl_2
ightarrow Hg_2Cl_2 \), which is the target reaction.

To find the \( \Delta H_{\text{rxn}} \), we add the \( \Delta H \) values of the two equations: \( \Delta H_{\text{rxn}} = -224 \, \text{kJ} + (-41 \, \text{kJ}) = -265 \, \text{kJ} \).

Answer:

\(-265\)