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.

the potential energy of the reactants is 200j.

from the energy diagram, the energy of the product formed is 350j; this means that, this reaction is an endothermic reaction, because it absorbs energy from its environment.

6.4 cm (max) - 25 cm (min)