Using the following thermochemical data: 2y(s) + 6hf(g) → 2yf3(s) + 3h2(g) δh° = –1811.0 kj/mol 2y(s) + 6hcl(g) → 2ycl3(s) + 3h2(g) δh° = –1446.2 kj/mol calculate δh° for the following reaction: yf3(s) + 3hcl(g) → ycl3(s) + 3hf(g).

ΔH° =   182.4 kJ/mol

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

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.

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explanation: