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# calculation Kc Watch

1. 3. Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) --> CH3COOH(g)

a) A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

for this I got value of Kc=1.53

c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted

and I got value of 2.91

and then it says using ur values is reaction exo or endo
2. (Original post by otrivine)
3. Ethanoic acid can be manufactured by the following reaction, which is carried out between 150 °C and 200 °C.
CH3OH(g) + CO(g) --> CH3COOH(g)

a) A mixture 17.6 moles of methanol and 19.6 moles of carbon monoxide is allowed to reach equilibrium at 175°C in a container with volume 5 dm3. It was found that 12.2 moles of ethanoic acid had been formed.

for this I got value of Kc=1.53

c) Another sample containing 17.6 moles of methanol and 19.6 moles of carbon monoxide was allowed to reach equilibrium, but at a lower temperature. This time it was found that 77.6% of methanol had reacted

and I got value of 2.91

and then it says using ur values is reaction exo or endo
If 77.6% has reacted = 0.776 * 17.6 = 13.66 hence remaining = 3.94 mol

The same moles of CO react hence CO remaining = 5.94

Moles of ethanoic acid = 13.66

Hence kc = 2.92

If your kc value is higher at a lower temperature the reaction is more to the RHS (side of the products) at a lower temperature. The forward reaction is exothermic.
3. (Original post by charco)

If your kc value is higher at a lower temperature the reaction is more to the RHS (side of the products) at a lower temperature. The forward reaction is exothermic.
However, is the calculation right, because the volume is 5dm3 as well for the second reaction, as I got 2.91?
4. (Original post by otrivine)
However, is the calculation right, because the volume is 5dm3 as well for the second reaction, as I got 2.91?
edited above ...
5. (Original post by charco)
edited above ...

I got 2.91 correct to 3 sig figures
i used the exact values when putting in Kc expression
6. (Original post by otrivine)
I got 2.91 correct to 3 sig figures
Near as makes no difference. I rounded to 2 dp.
7. (Original post by charco)
Near as makes no difference. I rounded to 2 dp.

Oh , I used the exact figures.

Lastly,a) MnO4- + U3+ → Mn2+ + UO2+

my overall balanced equation came to

H2O + 5/2 U3+ + Mno4- ---> 5/2UO2 + + 2H+ + Mn2+

is fine
8. (Original post by otrivine)
Oh , I used the exact figures.

Lastly,a) MnO4- + U3+ → Mn2+ + UO2+

my overall balanced equation came to

H2O + 5/2 U3+ + MnO4- ---> 5/2UO2 + + 2H+ + Mn2+

is fine
I should get rid of the fractions ...

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O
U3+ + 2H2O --> UO2+ + 4H+ + 2e

Multiply by 2 in equation 1 and by 5 in equation 2

2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O
5U3+ + 10H2O --> 5UO2+ + 20H+ + 10e
2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+

and cancel common terms

2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+
9. (Original post by charco)
I should get rid of the fractions ...

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O
U3+ + 2H2O --> UO2+ + 4H+ + 2e

Multiply by 2 in equation 1 and by 5 in equation 2

2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O
5U3+ + 10H2O --> 5UO2+ + 20H+ + 10e
2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+

and cancel common terms

2MnO4- + 5U3+ + 2H2O --> 2Mn2+ + 5UO2+ + 4H+
with fractions , do you lose marks?
10. (Original post by otrivine)
with fractions , do you lose marks?
That would depend entirely on the markscheme. To be on the safe side I would only ever use fractions in thermodynamics where you are restricted to mole quantities by the standard conditions.
11. (Original post by charco)
That would depend entirely on the markscheme. To be on the safe side I would only ever use fractions in thermodynamics where you are restricted to mole quantities by the standard conditions.
Ok will make sure thank you

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