Qualification > Sciences
ALL CIE CHEMISTRY DOUBTS HERE !!
Deadly_king:
--- Quote from: moon on October 17, 2010, 03:25:58 pm ---
I got the same answer -51.8 KJmol-1 but the markscheme's answer is +51.8KJmol-1 How come???
--- End quote ---
The answer is indeed +51.8 KJmol-1
Delta H = 2(-393.7) + 2(-285.9) - (-1411) = +51.8
2C + 2H2 ----> C2H4
Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O2 ----> CO2 (Enthalpy change of combustion of carbon or graphite)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of combustion of hydrogen)
3. C2H4 + 3O2 ----> 2CO2 + 2H2O (Enthalpy change of combustion of butane)
As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.
Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.
Let's say we'll convert 2 moles of carbon and 2 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to ethane.
In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.
Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of butane = 2(^C) + 2(^H) - (^C2H4)
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to butane whereas the enthalpy change of combustion of butane actually is the energy evolved when 1 mole of butane is completely burnt to carbon dioxide and water. In other words we are doing just the contrary. that's why we need to subtract it.
If you don't understand, let me know :)
$!$RatJumper$!$:
--- Quote from: moon on October 17, 2010, 03:25:58 pm ---
I got the same answer -51.8 KJmol-1 but the markscheme's answer is +51.8KJmol-1 How come???
--- End quote ---
You do get +51.8 when you do the calculation i mentioned
Deadly_king:
--- Quote from: $!$RatJumper$!$ on October 17, 2010, 03:41:29 pm ---
You do get +51.8 when you do the calculation i mentioned
--- End quote ---
Yeah.......you've done it correctly br0 :)
$!$RatJumper$!$:
--- Quote from: Deadly_king on October 17, 2010, 03:32:00 pm ---The answer is indeed +51.8 KJmol-1
Delta H = 2(-393.7) + 2(-285.9) - (-1411) = +51.8
2C + 2H2 ----> C2H4
Since the enthalpy changes of combustion has been given to you, you also need to write the equation relating each substance and its respective combustion.
1. C + O2 ----> CO2 (Enthalpy change of combustion of carbon or graphite)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of combustion of hydrogen)
3. C2H4 + 13/2 O2 ----> 2CO2 + 2H2O (Enthalpy change of combustion of butane)
As you may observe, the combustion of both hydrogen and carbon leads to carbon dioxide and hydrogen as well as the combustion of butane.
Hence, according to Hess's law, we may opt for another route in the conversion of carbon and hydrogen to butane.
Let's say we'll convert 2 moles of carbon and 2 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to ethane.
In practise these reactions are not feasible. But in theory we may opt for that method to find standard enthalpy changes of a particular reaction.
Let ^C be the enthalpy change of combustion of carbon.
So Enthalpy change of formation of butane = 2(^C) + 2(^H) - (^C2H4)
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to butane whereas the enthalpy change of combustion of butane actually is the energy evolved when 1 mole of butane is completely burnt to carbon dioxide and water. In other words we are doing just the contrary. that's why we need to subtract it.
If you don't understand, let me know :)
--- End quote ---
good one :)
moon:
I understood it now thanks Deadly_king and thanks 4 u too $!$RatJumper$!$
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