[SWalker92]

I have a few questions that I am stuck on and would really appreciate any help :)

1) Kp = 85 at 460°C and Kp = 21 at 800°C for the following reaction:

NO2(g) + SO2(g) <----> NO(g) + SO3(g)

Which one of the following statements is true?

a) △rH° > 0 and Kc < Kp
b) △rH° < 0 and Kc > Kp
c) △rH° = 0 and Kc = Kp
d) △rH° < 0 and Kc = Kp
e) △rH° > 0 and Kc = Kp

The answer says D but I can't figure out why.



2) Given the following thermochemical data, what is the enthalpy of formation △fH°298 for C6H6 (l) at 298K and 100kPa?

C6H6 (l) + 3H2(g) --> C6H12 (l) △H° = -206 kJ mol-1
6H2 (g) + 6C (s) --> C6H12 (l) △H° = -123 kJ mol-1


3) Samples of A (2.0 mol) and B (3.0 mol) are placed in a 10.0 L container and the following reaction takes place:

2A(g) <-----> 3B(g)

At equilibrium, the concentration of A is 0.14 M. What is the value of Kc?

[SWalker92]

Sorry, the  symbol represents a triangle :)

[Unknown008]

1. The is directly related to the yield. You know that Kc is a measure of the yield, do you? When the yield is greater, the delta H is more negative, because the reactants lose energy to get to a lower energy level (you can see that very clearly in a Born Haber Cycle).

Can you write down the Kc and Kp expressions? :)

2. First, write down the equation for the formation of C6H6:



You see that in the given equations, you can do this, by first passing through C6H12.



You started with 3 more H2 molecules, and you end up with 3 more H2 molecules... which means that you can take the given delta H without worrying about the excess H2 molecules.

3. If there are initially 2.0 mol of A and there is 0.14 mol of A left, how many mol of A got converted to B?

2 mol of A gives 3 mol of B. If you have the number of moles of A above which became B, how many mol of B did you get?

Using this now, use the equation for Kc.



The volume is kept constant, at 10, so you can find the final concentrations of A and B. Then, work Kc out! :)

Post what you get!