Answer :
To find the overall enthalpy for the reaction [tex]\( \text{P}_4\text{O}_6(s) + 2 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \)[/tex], we need to manipulate and combine the given reactions and their enthalpies. Here's the step-by-step process:
1. Given Chemical Reactions:
[tex]\[ \begin{array}{ll} \text{(1) } & \text{P}_4(s) + 3 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_6(s) \quad \Delta H_1 = -1640.1 \, \text{kJ} \\ \text{(2) } & \text{P}_4\text{O}_{10}(s) \rightarrow \text{P}_4(s) + 5 \text{O}_2(g) \quad \Delta H_2 = 2940.1 \, \text{kJ} \end{array} \][/tex]
2. Reverse the Second Reaction:
By reversing reaction (2), we get:
[tex]\[ \text{P}_4(s) + 5 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \][/tex]
Reversing the reaction changes the sign of the enthalpy change:
[tex]\[ \Delta H_2 = -2940.1 \, \text{kJ} \][/tex]
3. Combine the Reactions:
Now, add the modified reactions (1) and the reversed (2):
[tex]\[ \begin{array}{ll} \text{P}_4(s) + 3 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_6(s) & \Delta H_1 = -1640.1 \, \text{kJ} \\ \text{P}_4(s) + 5 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) & \Delta H_2 = -2940.1 \, \text{kJ} \end{array} \][/tex]
Simplify by canceling out [tex]\( \text{P}_4(s) \)[/tex] and combining [tex]\( \text{O}_2(g) \)[/tex]:
[tex]\[ \text{P}_4\text{O}_6(s) + 2 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \][/tex]
4. Calculate the Overall Enthalpy Change:
Add the enthalpy changes for the combined reactions:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 = -1640.1 \, \text{kJ} + (-2940.1 \, \text{kJ}) = -4580.2 \, \text{kJ} \][/tex]
5. Round to the Nearest Whole Number:
The nearest whole number to [tex]\(-4580.2 \, \text{kJ}\)[/tex] is [tex]\( -4580 \, \text{kJ} \)[/tex].
Therefore, the overall enthalpy for [tex]\( \text{P}_4\text{O}_6(s) + 2 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \)[/tex] is [tex]\( -4580 \, \text{kJ} \)[/tex].
1. Given Chemical Reactions:
[tex]\[ \begin{array}{ll} \text{(1) } & \text{P}_4(s) + 3 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_6(s) \quad \Delta H_1 = -1640.1 \, \text{kJ} \\ \text{(2) } & \text{P}_4\text{O}_{10}(s) \rightarrow \text{P}_4(s) + 5 \text{O}_2(g) \quad \Delta H_2 = 2940.1 \, \text{kJ} \end{array} \][/tex]
2. Reverse the Second Reaction:
By reversing reaction (2), we get:
[tex]\[ \text{P}_4(s) + 5 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \][/tex]
Reversing the reaction changes the sign of the enthalpy change:
[tex]\[ \Delta H_2 = -2940.1 \, \text{kJ} \][/tex]
3. Combine the Reactions:
Now, add the modified reactions (1) and the reversed (2):
[tex]\[ \begin{array}{ll} \text{P}_4(s) + 3 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_6(s) & \Delta H_1 = -1640.1 \, \text{kJ} \\ \text{P}_4(s) + 5 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) & \Delta H_2 = -2940.1 \, \text{kJ} \end{array} \][/tex]
Simplify by canceling out [tex]\( \text{P}_4(s) \)[/tex] and combining [tex]\( \text{O}_2(g) \)[/tex]:
[tex]\[ \text{P}_4\text{O}_6(s) + 2 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \][/tex]
4. Calculate the Overall Enthalpy Change:
Add the enthalpy changes for the combined reactions:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 = -1640.1 \, \text{kJ} + (-2940.1 \, \text{kJ}) = -4580.2 \, \text{kJ} \][/tex]
5. Round to the Nearest Whole Number:
The nearest whole number to [tex]\(-4580.2 \, \text{kJ}\)[/tex] is [tex]\( -4580 \, \text{kJ} \)[/tex].
Therefore, the overall enthalpy for [tex]\( \text{P}_4\text{O}_6(s) + 2 \text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \)[/tex] is [tex]\( -4580 \, \text{kJ} \)[/tex].
