Given: PCl5(s) → PCl3(g) + Cl2(g) ΔH°rxn = + 157 kJ
P4(g) + 6 Cl2(g) → 4 PCl3(g) ΔH°rxn = - 1207 kJ
What is the standard-state enthalpy change for the following reaction?
P4(g) + 10 Cl2(g) → 4PCl5(s)
a -2100 kJ
b-1835 kJ
c-1364 kJ
D -1786 kJ
Reverse the reaction-1 and multiply with 4
4 PCl3(g) + 4 Cl2(g) ------> 4 PCl5(s) delta H = -4*157 = -628 kJ
P4(g) + 6 Cl2(g) ? 4 PCl3(g) ?H°rxn = -1207 kJ
_------------------------------------------------------_
P4(g) + 10 Cl2(g) ? 4PCl5(s)
Delta H = -1207 kJ - 628 kJ = -1835 kJ
Thus, correct option is (b)
Use the standard reaction enthalpies given below to determine H°rxn for the following reaction: P4 (g) + 10 Cl2 (g) --> 4 PCl5 (s) ΔHreaction = ? Given: PCl5 (s) --> PCl3 (g) + Cl2 (g) ΔHreaction = +157 kJ P4 (g) + 6 Cl2 (g) --> 4 PCl3 (g) ΔHreaction = -1207 kJ Choices: A. -1835 kJ B. -1364 kJ C. -1050 kJ D. -1786 kJ E. -2100 kJ
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7. Use the reaction enthalpies given below to determine AHºrxn for the following reaction: 8. Use the AHºf information provided to calculate AHºrxn for the following: P4(g) + 10 C12(g) + 4PC15(s) AHørxn = ? SO2Cl2(g) + 2 H2O(l) → 2 HCI(g) + H2804 Given: PC15(s) - PC13(g) + Cl2(g) P4(g) + 6 Cl2(g) - 4 PC13(g) AHørxn- +157 kJ Hºrxn=-1207 kJ AHºf (kJ/mol) SO2Cl2(g) H2O(1) HCl(g) H2S040) -364 -286 -92 -814 Voboty A) -2100. kJ B) -1835 kJ C)...
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