Qualification > Sciences
who have answers to this book???(edexcel)
~ Miss Relina ~:
me agaun that my questions and for question concerning calorimeter i have attatched the picture for (fig1.2.10)
Deadly_king:
--- Quote from: Relina on December 07, 2010, 10:21:05 pm ---me agaun that my questions and for question concerning calorimeter i have attatched the picture for (fig1.2.10)
--- End quote ---
1st attachment. No 3
is exothermic since we can note from the graph that the products are at a lower temperature than the reactants. This means that they possess less energy. Therefore energy has been released during the reaction which demonstrates that the reaction is indeed exothermic.
is endothermic. Same principle as above is applied. I guess you'll be able to put n your own words. ;)
Temperature change for A - exothermic : (30 - 10) = 20oC
Temperature change for B - endothermic : (40 - 20) = 20oC
4th attachment
You should use the two equations provided below to form the required equation which is :
H2SO4 + 2KCl -----> 2HCl + K2SO4
Now we need to re-arrange the information provided to get this. You need to inverse the 2nd equation provided and multiply it by 2.
H2SO4 + 2KOH ----> K2SO4 + 2H2O ^H = -324 KJ
2KCl + 2H2O ----> 2HCl + 2KOH ^H = +408 KJ
NOTE : When you inverse an equation and multiply it by 2, you should inverse the enthalpy change as well and multiply it too.
Now you mix the two equations.
H2SO4 + 2KOH + 2KCl + 2H2O ----> K2SO4 + 2H2O + 2HCl + 2KOH
From this resulting equation you can note that 2KOH and 2H2O cancel each other from both sides of the equation.
So now you get your equation. For its enthalpy change you just need to add them tegether.
^ H = -324 + 408 = +84 KJ
Deadly_king:
5th attachment
This is the equation representing the formation of propanoic acid.
3C + 3H2 + 1/2O2 ----> C3H5OH
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 formation of carbon dioxide)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of formation of water)
3. C3H5OH + 9/2O2 ----> 3CO2 + 3H2O (Enthalpy change of combustion of propanoic acid)
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 3 moles of carbon and 3 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to propanoic acid.
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 propanoic acid = 3(^C) + 3(^H) - (^C3H5OH) = 3(-393.5) + 3(-285.8) - (-1527.2) = -510.7 KJ
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to propanoic acid whereas the enthalpy change of combustion of propanoic acid actually is the energy evolved when 1 mole of propanoic acid 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.
~ Miss Relina ~:
--- Quote from: Martin Luther King Jr on December 09, 2010, 12:42:04 pm ---5th attachment
This is the equation representing the formation of propanoic acid.
3C + 3H2 + 1/2O2 ----> C3H5OH
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 formation of carbon dioxide)
2. H2 + 0.5O2 ----> H2O (Enthalpy change of formation of water)
3. C3H5OH + 9/2O2 ----> 3CO2 + 3H2O (Enthalpy change of combustion of propanoic acid)
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 3 moles of carbon and 3 moles of Hydrogen gas to carbon dioxide and water. Then these two products are converted to propanoic acid.
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 propanoic acid = 3(^C) + 3(^H) - (^C3H5OH) = 3(-393.5) + 3(-285.8) - (-1527.2) = -510.7 KJ
NOTE : We need to subtract the enthalpy change of combustion since we are converting carbon dioxide and water back to propanoic acid whereas the enthalpy change of combustion of propanoic acid actually is the energy evolved when 1 mole of propanoic acid 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.
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Thanks alot I'll post other questions later
Deadly_king:
--- Quote from: Relina on December 16, 2010, 11:01:59 pm ---Thanks alot I'll post other questions later
--- End quote ---
You're welcome. :D
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