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Consider the following intermediate chemical equations:

[tex]\[
\begin{array}{ll}
P_4(s) + 6Cl_2(g) \rightarrow 4PCl_3(g) & \Delta H_1 = -2,439 \, \text{kJ} \\
4PCl_5(g) \rightarrow P_4(s) + 10Cl_2(g) & \Delta H_2 = 3,438 \, \text{kJ}
\end{array}
\][/tex]

What is the enthalpy of the overall chemical reaction [tex]\( PCl_3(g) \rightarrow PCl_5(g) + Cl_2(g) \)[/tex]?

A. [tex]\(-999 \, \text{kJ}\)[/tex]
B. [tex]\(-250 \, \text{kJ}\)[/tex]
C. [tex]\(250 \, \text{kJ}\)[/tex]



Answer :

GinnyAnswer
To determine the enthalpy change of the overall chemical reaction, we need to consider the enthalpies of the given intermediate reactions and sum them up accordingly.

The given intermediate chemical equations along with their enthalpy changes are:
[tex]\[P_4(s) + 6 \text{Cl}_2(g) \rightarrow 4 \text{PCl}_2(g) \quad \Delta H_1 = -2439 \text{kJ}\][/tex]
[tex]\[4 \text{PCl}_5(g) \rightarrow P_4(s) + 10 \text{Cl}_2(g) \quad \Delta H_2 = 3438 \text{kJ}\][/tex]

1. [tex]\(\Delta H_1 = -2439 \text{kJ}\)[/tex]
2. [tex]\(\Delta H_2 = 3438 \text{kJ}\)[/tex]

To find the enthalpy change of the overall reaction, we sum the enthalpy changes of the intermediate reactions:

[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 \][/tex]

Substituting the given values:

[tex]\[ \Delta H_{\text{overall}} = (-2439 \text{kJ}) + (3438 \text{kJ}) = 999 \text{kJ} \][/tex]

Therefore, the enthalpy change for the overall chemical reaction is [tex]\(\Delta H_{\text{overall}} = 999 \text{kJ}\)[/tex].

Given the options:
- [tex]\(-999 \text{kJ}\)[/tex]
- [tex]\(-250 \text{kJ}\)[/tex]
- [tex]\(250 \text{kJ}\)[/tex]

The enthalpy change of the overall chemical reaction does not match any of the provided options directly, but considering the calculation, the closest option by magnitude is neither of the negative values nor 250 kJ directly.

Thus the enthalpy change for the reaction [tex]$PCl_1(g) \rightarrow PCl_1(g) + Cl_2(g)$[/tex] with the provided context is 999 kJ.

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