#### Question

20. This diagram would represent the enthalpy changes in which of the following?

Refer to Picture http://tinypic.com/r/5p515l/7

hot pack

boiling liquid

cold pack

melting solid

19. What is the change in enthalpy when 250 g of steam condenses at 100 C? (ΔHv = 40.67 kJ/mol)

(Points : 3)

-565 kJ

-291 kJ

291 kJ

565 kJ

Question one:

It is the hot pack.

That's because it involves reactants and products, so must be a chemical reaction. The boiling liquid and melting solid are just phase changes. They're not chemical reactions, just physical changes.

It isn't the cold pack because as seen from the diagram the energy level of the reactants is higher than the products. Energy is given out as the reaction progresses. If the energy is given out as heat, it would be exothermic thus can be a hot pack but not a cold pack.

Question two:

ΔH of vaporization = 40.67

therefore, ΔH of condensation = -40.67 kJ/mol

Change in enthalpy

= ΔHcondensation x number of moles of steam (water)

= ΔHcondensation x (mass of water / RMM of water)

= -40.67 x (250/18)

= -564.8611111 kJ

= -565 kJ (3 significant figures)

#1

it is exothermic - if by hot pack you mean a rnx that heats up then that one

250 g/18 g/mole * 40.67 = 564.9 kJ

since it is a condensation - energy is released

- 565 kJ

#2

1. Hot Pack

19. n = m/M M is the molar mass of h20

n = 250g (18)

n = 13.889 moles

13.8889 * (-40.67) = -565 kJ

It is negative as it is condensation -- The positive value given is heat of vaporization

#3