Question

A studen is asket to calculate the amount of heat involved in changing 10.0 g of liquid bomine at room temperature (22.5 C) to vapor at 59.0 C. To do this, one must know specific heat (0.474 J/g*C), boiling point (59 C), and heat of vaporization (29.6 kJ/mol) of bromine.In addition, the following step-wise process must be followed.

a) Calculate delta H for: Br2 (l, 22.5 C) --> Br2 (l, 59.0 C)

b) Calculate delta H for: Br2 (l, 59.0 C) --> Br2 (g, 59.0 C)

c) Using Hess's law, calculate delta H for: Br2 (l, 22.5 C) --> Br2 (g, 59.0 C)

Answer 1

You are given the steps to follow. What's the trouble?

Answer 2

I don't understand how you do (b), I got 173kJ for (a) and I understand Hess's law for (c), but (b) I really don't understand (b)

Answer 3

Since its a phase change in part b you have to calculate the delta h vap so in order to do that convert the 10 grams to mols then convert that to kj. ( delta h vap for one mole of br2= 29.6 kj). So 10 g / 159.80 x 29.6 kj = 1.85 kj

Answer 4

What about the c part