I got -158 kJ Given the following thermochemical equations, HF(g) --> H(g) + F(g
ID: 691194 • Letter: I
Question
I got -158 kJ
Given the following thermochemical equations,
HF(g) --> H(g) + F(g)
568 kJ (per mol HF)
H2(g) --> 2 H(g)
436 kJ (per mol H2)
HF(g) --> (1/2) H2(g) + (1/2) F2(g)
271 kJ (per mol HF)
what is the standard enthalpy change for the reaction below?
2 F(g) --> F2(g)
Give your answer in kJ (per mol F2), accurate tothree significant figures.
HF(g) --> H(g) + F(g)
568 kJ (per mol HF)
H2(g) --> 2 H(g)
436 kJ (per mol H2)
HF(g) --> (1/2) H2(g) + (1/2) F2(g)
271 kJ (per mol HF)
Explanation / Answer
HF(g) --> H(g) + F(g)
H1= 568 kJ (per mol HF) ----(1)
H2(g) --> 2 H(g)
H2=436 kJ (per mol H2) ----(2)
HF(g) --> (1/2) H2(g) + (1/2) F2(g)
H3=271 kJ (per mol HF) ---(3)
HF(g) --> H(g) + F(g)
H1= 568 kJ (per mol HF) ----(1)
H2(g) --> 2 H(g)
H2=436 kJ (per mol H2) ----(2)
HF(g) --> (1/2) H2(g) + (1/2) F2(g)
H3=271 kJ (per mol HF) ---(3)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.