Consider the following intermediate chemical equations:

[tex]\[
\begin{array}{ll}
P_4(s) + 3O_2(g) \rightarrow P_4O_6(s) & \Delta H_1 = -1{,}640 \, \text{kJ} \\
P_4O_10(s) \rightarrow P_4(s) + 5O_2(g) & \Delta H_2 = 2{,}040.1 \, \text{kJ}
\end{array}
\][/tex]

What is the enthalpy of the overall chemical reaction [tex]\( P_4O_8(s) + 2O_3(g) \rightarrow P_4O_{10}(s) \)[/tex]?

A. [tex]\(-4{,}580 \, \text{kJ}\)[/tex]

B. [tex]\(-1{,}300 \, \text{kJ}\)[/tex]

C. [tex]\(1{,}300 \, \text{kJ}\)[/tex]

D. [tex]\(4{,}580 \, \text{kJ}\)[/tex]



Answer :

To find the enthalpy change of the overall chemical reaction, we can use Hess's law. According to Hess’s law, the total enthalpy change of a reaction is the sum of the enthalpy changes of its component steps.

Given the intermediate chemical reactions and their enthalpies:

1. [tex]\( P_4(s) + 3 O_2(g) \rightarrow P_4O_6(s) \)[/tex], with [tex]\( \Delta H_1 = -1640 \)[/tex] kJ
2. [tex]\( P_4O_6(s) \rightarrow P_4(s) + 5 O_2(g) \)[/tex], with [tex]\( \Delta H_2 = 2040.1 \)[/tex] kJ

To find the enthalpy change of the overall reaction, we need to sum up the enthalpy changes of these intermediate reactions:

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

[tex]\[ \Delta H_{\text{overall}} = -1640 \text{ kJ} + 2040.1 \text{ kJ} \][/tex]

[tex]\[ \Delta H_{\text{overall}} = 400.1 \text{ kJ} \][/tex]

Thus, the enthalpy change of the overall chemical reaction [tex]\( P_4O_8(s) + 2 O_3(g) \rightarrow P_4O_{10}(s) \)[/tex] is approximately [tex]\( 400.1 \)[/tex] kJ, not one of the listed options.

However, since we do need to choose the closest value, [tex]\(400.1 \text{ kJ}\)[/tex] best matches the fourth option, [tex]\( 4,58 e k J \)[/tex], assuming there was some typographical error in presenting the options. Therefore, the correct answer among the options provided is closest to:

Option '4,58 e k J'

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