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Sagot :
To determine the enthalpy change for the overall reaction [tex]\(P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\)[/tex], we can use the provided enthalpy changes for the given reactions and the concept of reversing reactions and adding them together.
Firstly, let's consider the two given reactions and their enthalpy changes:
1. [tex]\(P_4(s) + 3O_2(g) \rightarrow P_4O_6(s)\)[/tex] with [tex]\(\Delta H_1 = -1640 \, \text{kJ}\)[/tex]
2. [tex]\(P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g)\)[/tex] with [tex]\(\Delta H_2 = 2840.1 \, \text{kJ}\)[/tex]
Since we need the overall reaction [tex]\(P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\)[/tex], we need to manipulate the given reactions to result in this overall reaction.
Step 1: Reverse the first reaction
Reversing the given first reaction [tex]\(P_4(s) + 3 O_2(g) \rightarrow P_4O_6(s)\)[/tex] will give us:
[tex]\[ P_4O_6(s) \rightarrow P_4(s) + 3 O_2(g) \][/tex]
Reversing a reaction changes the sign of [tex]\(\Delta H_1\)[/tex]. Thus, the enthalpy change for the reversed reaction is:
[tex]\[ \Delta H_{\text{reversed}} = 1640 \, \text{kJ} \][/tex]
Step 2: Combine the reversed first reaction with the second reaction
We now have:
[tex]\[ P_4O_6(s) \rightarrow P_4(s) + 3O_2(g) \quad \text{with} \quad \Delta H = 1640 \, \text{kJ} \][/tex]
[tex]\[ P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g) \quad \text{with} \quad \Delta H = 2840.1 \, \text{kJ} \][/tex]
Considering these together:
[tex]\[ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \quad \Delta H_2 = 2840.1 \, \text{kJ} \][/tex]
Now, we need the overall target reaction:
[tex]\[ P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
If we combine these equations appropriately, we subtract [tex]\(P_4(s) + 3O_2(g)\)[/tex].
Step 3: Add the enthalpy changes
For the overall target reaction:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_{\text{reversed}} + \Delta H_2 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 1640 \, \text{kJ} + 2840.1 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 4480.1 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the overall reaction [tex]\(P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\)[/tex] is [tex]\(4480.1 \, \text{kJ}\)[/tex].
The correct answer from the given options is [tex]\(4480.1 \, \text{kJ}\)[/tex], which matches the closest to:
[tex]\[ \boxed{4,580 \, \text{kJ}} \][/tex]
Firstly, let's consider the two given reactions and their enthalpy changes:
1. [tex]\(P_4(s) + 3O_2(g) \rightarrow P_4O_6(s)\)[/tex] with [tex]\(\Delta H_1 = -1640 \, \text{kJ}\)[/tex]
2. [tex]\(P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g)\)[/tex] with [tex]\(\Delta H_2 = 2840.1 \, \text{kJ}\)[/tex]
Since we need the overall reaction [tex]\(P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\)[/tex], we need to manipulate the given reactions to result in this overall reaction.
Step 1: Reverse the first reaction
Reversing the given first reaction [tex]\(P_4(s) + 3 O_2(g) \rightarrow P_4O_6(s)\)[/tex] will give us:
[tex]\[ P_4O_6(s) \rightarrow P_4(s) + 3 O_2(g) \][/tex]
Reversing a reaction changes the sign of [tex]\(\Delta H_1\)[/tex]. Thus, the enthalpy change for the reversed reaction is:
[tex]\[ \Delta H_{\text{reversed}} = 1640 \, \text{kJ} \][/tex]
Step 2: Combine the reversed first reaction with the second reaction
We now have:
[tex]\[ P_4O_6(s) \rightarrow P_4(s) + 3O_2(g) \quad \text{with} \quad \Delta H = 1640 \, \text{kJ} \][/tex]
[tex]\[ P_4O_{10}(s) \rightarrow P_4(s) + 5O_2(g) \quad \text{with} \quad \Delta H = 2840.1 \, \text{kJ} \][/tex]
Considering these together:
[tex]\[ P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s) \quad \Delta H_2 = 2840.1 \, \text{kJ} \][/tex]
Now, we need the overall target reaction:
[tex]\[ P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s) \][/tex]
If we combine these equations appropriately, we subtract [tex]\(P_4(s) + 3O_2(g)\)[/tex].
Step 3: Add the enthalpy changes
For the overall target reaction:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_{\text{reversed}} + \Delta H_2 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 1640 \, \text{kJ} + 2840.1 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 4480.1 \, \text{kJ} \][/tex]
Thus, the enthalpy change for the overall reaction [tex]\(P_4O_8(s) + 2O_2(g) \rightarrow P_4O_{10}(s)\)[/tex] is [tex]\(4480.1 \, \text{kJ}\)[/tex].
The correct answer from the given options is [tex]\(4480.1 \, \text{kJ}\)[/tex], which matches the closest to:
[tex]\[ \boxed{4,580 \, \text{kJ}} \][/tex]
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