(Please include the working to reach your correct response)
The Keq for the equilibrium below is 0.112 at 700 °C.
SO2 (g) + 1/2 O2 (g) <----->SO3 (g)
What is the value of Keq at this temperature for the following reaction.
2SO3 (g) <-----> 2SO2 (g) + O2 (g)
a. 79.7
b. 2.99
c. 17.86
d. 4.46
e. 8.93
2007-03-12 08:53:18 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Since Keq=concentration of products/reactants
SO2+1/2O2 <->SO3
.112=[SO3]/[SO2][O2]^1/2
Since you are looking for:
2SO3<->2SO2+O2
it is the ricipricol of .112 because you have changed the products and reactants. so now the products of the reation from above will go below and the reactants will become products
Also, since the coefficents in the 2nd equation are doubled, you have to square all the concentration
so,
1/.112=8.93
then squaring it will give you: 79.7
the answer is A
2007-03-12 09:23:28 · answer #1 · answered by blueboy3056 3 · 4⤊ 4⤋
(a). lowering the quantity skill increasing the rigidity. LeChatelier's concept says that the device will respond by way of lowering the rigidity, meaning reducing the form of molecules contemporary, and the reaction will shift to the left.
2016-10-02 00:21:45 · answer #2 · answered by Anonymous · 0⤊ 1⤋
You've turned the equation backwards, and doubled it.
So you have to take the reciprocal of 0.112, and then square it.
a is the answer.
2007-03-12 09:07:07 · answer #3 · answered by Gervald F 7 · 6⤊ 2⤋
.0125 is the correct Ans.
2014-04-04 01:09:35 · answer #4 · answered by Arshin 2 · 2⤊ 0⤋