QuestionApril 30, 2025

76. Given the following data P_(4)(s)+6Cl_(2)(g)arrow 4PCl_(3)(g) Delta H=-1225.6kJ P_(4)(s)+5O_(2)(g)arrow P_(4)O_(10)(s) Delta H=-2967.3kJ PCl_(3)(g)+Cl_(2)(g)arrow PCl_(5)(g) Delta H=-84.2kJ PCl_(3)(g)+(1)/(2)O_(2)(g)arrow Cl_(3)PO(g) Delta H=-285.7kJ calculate Delta H for the reaction P_(4)O_(10)(s)+6PCl_(5)(g)arrow 10Cl_(3)PO(g)

76. Given the following data P_(4)(s)+6Cl_(2)(g)arrow 4PCl_(3)(g) Delta H=-1225.6kJ P_(4)(s)+5O_(2)(g)arrow P_(4)O_(10)(s) Delta H=-2967.3kJ PCl_(3)(g)+Cl_(2)(g)arrow PCl_(5)(g) Delta H=-84.2kJ PCl_(3)(g)+(1)/(2)O_(2)(g)arrow Cl_(3)PO(g) Delta H=-285.7kJ calculate Delta H for the reaction P_(4)O_(10)(s)+6PCl_(5)(g)arrow 10Cl_(3)PO(g)
76. Given the following data P_(4)(s)+6Cl_(2)(g)arrow 4PCl_(3)(g) Delta H=-1225.6kJ P_(4)(s)+5O_(2)(g)arrow P_(4)O_(10)(s) Delta H=-2967.3kJ PCl_(3)(g)+Cl_(2)(g)arrow PCl_(5)(g) Delta H=-84.2kJ PCl_(3)(g)+(1)/(2)O_(2)(g)arrow Cl_(3)PO(g) Delta H=-285.7kJ calculate Delta H for the reaction P_(4)O_(10)(s)+6PCl_(5)(g)arrow 10Cl_(3)PO(g)

Solution
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Answer

\Delta H = -5104.9 kJ Explanation 1. Write the target reaction P_{4}O_{10}(s) + 6PCl_{5}(g) \rightarrow 10Cl_{3}PO(g) 2. Reverse and scale reactions Reverse the first reaction: 4PCl_{3}(g) \rightarrow P_{4}(s) + 6Cl_{2}(g), \Delta H = +1225.6 kJ. Scale the second reaction by 1: P_{4}(s) + 5O_{2}(g) \rightarrow P_{4}O_{10}(s), \Delta H = -2967.3 kJ. Scale the third reaction by 6: 6PCl_{3}(g) + 6Cl_{2}(g) \rightarrow 6PCl_{5}(g), \Delta H = 6(-84.2) kJ. Scale the fourth reaction by 10: 10PCl_{3}(g) + 5O_{2}(g) \rightarrow 10Cl_{3}PO(g), \Delta H = 10(-285.7) kJ. 3. Sum the enthalpies Add the enthalpies of the modified reactions: \Delta H = (+1225.6) + (-2967.3) + 6(-84.2) + 10(-285.7) 4. Calculate the total enthalpy change \Delta H = 1225.6 - 2967.3 - 505.2 - 2857

Explanation

1. Write the target reaction<br /> $P_{4}O_{10}(s) + 6PCl_{5}(g) \rightarrow 10Cl_{3}PO(g)$<br /><br />2. Reverse and scale reactions<br /> Reverse the first reaction: $4PCl_{3}(g) \rightarrow P_{4}(s) + 6Cl_{2}(g)$, $\Delta H = +1225.6$ kJ.<br /> Scale the second reaction by 1: $P_{4}(s) + 5O_{2}(g) \rightarrow P_{4}O_{10}(s)$, $\Delta H = -2967.3$ kJ.<br /> Scale the third reaction by 6: $6PCl_{3}(g) + 6Cl_{2}(g) \rightarrow 6PCl_{5}(g)$, $\Delta H = 6(-84.2)$ kJ.<br /> Scale the fourth reaction by 10: $10PCl_{3}(g) + 5O_{2}(g) \rightarrow 10Cl_{3}PO(g)$, $\Delta H = 10(-285.7)$ kJ.<br /><br />3. Sum the enthalpies<br /> Add the enthalpies of the modified reactions: <br /> $\Delta H = (+1225.6) + (-2967.3) + 6(-84.2) + 10(-285.7)$<br /><br />4. Calculate the total enthalpy change<br /> $\Delta H = 1225.6 - 2967.3 - 505.2 - 2857$
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