Answer:
ΔH° = 182.4 kJ/mol
Step-by-step explanation:
The ΔH wanted is for the reaction :
YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g)
This is a Hess Law problem where e will have to algebraically manipulate the first and second equations , add them together, and arrive at the desired equation above.
Notice if we reverse the first equation and divide it by 2 and add to the the second only divided by two, we will arrive to the desired equation:
2YF3(s) + 3H2(g) → 2Y(s) + 6HF(g) ΔH° = 1811.0 kJ/mol (change sign)
dividing by two :
YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1
2Y(s) + 6HCl(g) → 2YCl3(s) + 3H2(g) ΔH° = –1446.2 kJ/mol
dividing this one by two,
Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2
Now adding 1 and 2
YF3(s) + 3/2H2(g) → Y(s) + 3HF(g) ΔH° = 905.5 kJ/mol Eq 1
Y(s) + 3HCl(g) → YCl3(s) + 3/2 H2(g) ΔH° = –1446.2 kJ/mol/2 = - 723.10 kJ/mol Eq 2
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YF3(s) + 3HCl(g) → YCl3(s) + 3HF(g). ΔH° = 905.5 + (-723.1) kJ/mol
ΔH° = 182.4 kJ/mol
Notice how the Y(s) and H2 cancel nicely and the coefficients are the right ones.