In: Chemistry
Given the enthalpies of reaction: 2P(g) + 3Cl2(g) → 2PCl3(g) DH
= –574 kJ 2P(g) + 5Cl2(g) → 2PCl5(g) DH = –887 kJ What is the
enthalpy change of the following reaction: PCl3(g) + Cl2(g) →
PCl5(g)
Question 5 options:
1461 kJ
–1461 kJ
222 kJ
–313 kJ
–156 kJ
2P(g) + 3Cl2(g) → 2PCl3(g) DH = –574 kJ
2P(g) + 5Cl2(g) → 2PCl5(g) DH = –887 kJ
(-) (-) (-) (+)
------------------------------------------------------------------
2PCl5(g) -----------> 2PCl3(g) + 2Cl2(g) DH = 313KJ
PCl5(g) -----------> PCl3(g) + Cl2(g) DH = 313/2 = 156.5KJ
PCl3(g) + Cl2(g) ---------------> PCl5(g) DH = -156.5KJ >>>>>answer
–156 kJ >>>>answer