jackinthepatch's chemistry questions

[jackinthepatch]

Given that the equilibrium constant for the reaction
            2A + 2B 3C + D
has a value of 10, then the equilibrium constant for the reaction
            6C + 2D 4A + 4B
would be:

a. 0.10
b. 0.010
c. 100
d. 0.050

I desperately need someone to explain how you get the answer lol. Thanks

[/0]

(squaring)

(reciprocating)

[jackinthepatch]

Ah thanks

[jackinthepatch]

Fe³⁺ ions readily react with SCN^- ions according to the equation:

      Fe³⁺ + SCN^- Fe(SCN)²⁺

When 0.060 mol of Fe³⁺ and 0.13 mol of SCN^- are added to 0.50L of water, the resulting solution is found to contain 0.030 mol of Fe(SCN)²⁺ at equilibrium. Calculate a value for the equilibium constant.


I think I know the answer...but then again, I can never be sure lol. Thanks.

[/0]

Not 100% about this, hope it's the same answer you got

[kamil9876]

We need the concentrations of the and at equilibrium(hence the number of moles after reaction). If 0.03mol of FeSCN^{2+} was created then 0.03mol of and must've been used up as shown by the equation. (1:1:1 ratio on the coefficients)

hence


Now from here you can find the concentration of these. and respectively. While the concentration of is . Now you can plug in.

[jackinthepatch]

Yeah that's what I got, thanks for that guys, sometimes ya need the reassurance haha.

[jackinthepatch]

Again I think I know the answer but just wanna check:

0.0500 mol of SO₂, 0.0100 mol of O₂ and 0.150 mol of SO₃ were mixed in a 2.00L vessel and allowed to reach equilibrium according to this equation:

        2SO₂ + O₂ 2SO₃

Analysis showed that 0.140 mol of SO₃ remained. Calculate K at this temperature.

[NE2000]

Again I think I know the answer but just wanna check:

0.0500 mol of SO₂, 0.0100 mol of O₂ and 0.150 mol of SO₃ were mixed in a 2.00L vessel and allowed to reach equilibrium according to this equation:

        2SO₂ + O₂ 2SO₃

Analysis showed that 0.140 mol of SO₃ remained. Calculate K at this temperature.

0.140 mol of sulfur trioxide remained. Therefore 0.010 mol of sulfur trioxide was used. Therefore 0.010 mol of sulfur dioxide was used and 0.0050 mol of oxygen gas was used. The result is at equilibrium:

[SO3]=0.140/2=0.070 M
[SO2]=0.060/2=0.030 M
[O2] = 0.015/2=0.0075 M

Sub values in to the equilibrium equation to get around 726 .

[jackinthepatch]

Ah thanks mate.

Also, how do you represent rate of reaction in terms of units/values?
For example, if a reaction resulted in the release of CO₂ over a period of time, how would the rate of reaction be represented?

[mark_alec]

mol/second, mol/minute, mol/hour or whatever is an appropriate unit for the amount of stuff (number of particles? moles? millimoles?) and time period.