Question

0.200 mol of sulfur dioxide and 0.200 mol of oxygen are placed in a 1.00 dm3 sealed container. The gases are allowed to react until equilibrium is reached.

2SO2 + O2 = 2SO3

At equilibrium there is 0.100 mol of SO3 in the container.
What is the value of Kc?
A 0.150 mol dm–3
B 0.800 mol dm–3
C 1.25 mol–1 dm3
D 6.67 mol–1 dm3
( the answer is D) but i don't understand how , please explain method...

Answer 1

Answer:

option C is correct....must be wrong printed answer

Explanation:

Answer 2

Answer:

So the answer is D

Explanation

They already give you the moles at equilibrium of sulphur trioxide so you first must calculate the moles at equilibrium for the reactants.

For sulphur dioxide you do 0.200 mol minus 0.100 mol because its a ratio of 2 to 2 and to find the moles at equilibrium you must do the initial moles given to you minus the moles used up in the reaction (0.100), for oxygen you do 0.200 minus 0.150 mol due to the ratio of 2 to 1 you do 0.200 mol minus 0.05 mol.

So you have for SO2 0.100 mol and for 02 you have 0.150 mol.

Then you use the formula for Kc which is products over reactants, you do 0.1 squared over 0.1 squared times 0.15 which should give you 0.666...

which you round up to 6.67. Then you find the units which is a mol dm3 over 2 mol dm3 giving you mol-1 dm3.

Hope this helped.