Question

Enthalpy
I need step-by-step guidance please and thank you in advance.
Question Below:

Answer 1

@Jhannybean

Answer 2

\[\Delta H = \sum {n\cdot \text{products} -\sum m \cdot \text{reactants}}\] m and n are coefficients of the reactants and products,respectively.

Answer 3

So you're given \(\Delta H = -89 \frac{kJ}{mol}\)

Answer 4

yeah i think I manage it now,
just to make sure Hf for I2 is zero right?

Answer 5

**can

Answer 6

Yes, all diatomics and elements

Answer 7

thanks!

Answer 8

\[-89 \frac{kJ}{mol} = -840\frac{kJ}{mol} +2\text{IF} -\left( -942 \frac{kJ}{mol}\right)\]

Answer 9

that's what i did and i got -94.5 kJ

Answer 10

- 89 = 2x - 840 + 942

-89 = 2x +102

-89 - 102 = 2x

-191 = 2x

x = -191/2 = -95.5