Question

Ung the table below, what is the change in enthalpy for the following reaction?
I2 (s) -> I2 (g)
Substance ΔHf(kJ/mol)
I2 (g) 62.4
I2 (s) 0.0
Answer

−62.4 kJ

6.24 x 104 J

−6.24 x 104 J

0 J

Answer 1

@ganeshie8

Answer 2

@nincompoop

Answer 3

@aaronq

Answer 4

Subtract the enthalpy of the reactants from the products

\(\Delta H=H_{Products} - H_{reactants}\)

Answer 5

so...-62.4?

Answer 6

no, dH=62.4 kJ -0 kJ

Answer 7

sorry, this is probably a stupid question, but what is dH?

Answer 8

the products?

Answer 9

\(\Delta H=dH\)

Answer 10

you'll see it written both ways

Answer 11

ah...ok...
6.24 x 104 J?

Answer 12

yep

Answer 13

gratzi! :D

Answer 14

np!