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user342
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#1
https://isaacphysics.org/questions/p...c-8021925d7d27
Not sure how to go about part C
Not sure how to go about part C
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charco
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#2
(Original post by user342)
https://isaacphysics.org/questions/p...c-8021925d7d27
Not sure how to go about part C
https://isaacphysics.org/questions/p...c-8021925d7d27
Not sure how to go about part C
Draw a route map from P4(s) + nO2(g) ====> gaseous atoms ==> P4On(s)
Taking into account all of the energy steps needed
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#3
(Original post by charco)
You need the formula of the phosphorus oxide from earlier, then:
Draw a route map from P4(s) + nO2(g) ====> gaseous atoms ==> P4On(s)
Taking into account all of the energy steps needed
You need the formula of the phosphorus oxide from earlier, then:
Draw a route map from P4(s) + nO2(g) ====> gaseous atoms ==> P4On(s)
Taking into account all of the energy steps needed
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#4
(Original post by user342)
Ok so it would be energy to form P4O10 subtract energy to vaporise P4 and then plus (or minus?) energy to vaporise P4O10... I got energy to form P4O10 as 6352 (12x376 for the single P-O bonds + 4x460 for the double) but it's telling me the answer I've gotten for this method is wrong. Could you nudge me in the right direction again?
Ok so it would be energy to form P4O10 subtract energy to vaporise P4 and then plus (or minus?) energy to vaporise P4O10... I got energy to form P4O10 as 6352 (12x376 for the single P-O bonds + 4x460 for the double) but it's telling me the answer I've gotten for this method is wrong. Could you nudge me in the right direction again?
The enthalpy of vaporization of P4 is 51.9 kJ mol−1 and that of P4O10 is 93.52kJ mol−1.
Using bond enthalpy date you must explode the reactants into gaseous atoms and then reform them as the compound:
P4(s) + 5O2(g) ==> P4(g) + 5O2(g) ==> 4P(g) + 10O(g) ==> P4O10(g) ==> P4O10(s)
Then add up the energy for each transformation.
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#5
(Original post by charco)
OK, you have the formula correct for the oxide, P4O10
The enthalpy of vaporization of P4 is 51.9 kJ mol−1 and that of P4O10 is 93.52kJ mol−1.
Using bond enthalpy date you must explode the reactants into gaseous atoms and then reform them as the compound:
P4(s) + 5O2(g) ==> P4(g) + 5O2(g) ==> 4P(g) + 10O(g) ==> P4O10(g) ==> P4O10(s)
Then add up the energy for each transformation.
OK, you have the formula correct for the oxide, P4O10
The enthalpy of vaporization of P4 is 51.9 kJ mol−1 and that of P4O10 is 93.52kJ mol−1.
Using bond enthalpy date you must explode the reactants into gaseous atoms and then reform them as the compound:
P4(s) + 5O2(g) ==> P4(g) + 5O2(g) ==> 4P(g) + 10O(g) ==> P4O10(g) ==> P4O10(s)
Then add up the energy for each transformation.
From the hints I got that it would be: 93.52 (energy for products) - (51.9+6352) (energy of reactants), which gives -6310, but it says that's incorrect. Where am I going wrong?
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charco
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#6
(Original post by user342)
I'm still confused
From the hints I got that it would be: 93.52 (energy for products) - (51.9+6352) (energy of reactants), which gives -6310, but it says that's incorrect. Where am I going wrong?
I'm still confused
From the hints I got that it would be: 93.52 (energy for products) - (51.9+6352) (energy of reactants), which gives -6310, but it says that's incorrect. Where am I going wrong?
STEP 2: You then have to convert P4(g) into 4P(g) - this is six times the bond enthalpy of phosphorus (look at the diagram of the molecule) = 6 x 201 = +1206 kJ
STEP 3: Now you need to convert 5O2(g) into 10O(g) - this is five times the bond enthalpy of oxygen = 5 x 495 kJ = +2475 kJ
STEP 4: Now we have to assemble the molecule, which consists of 12 x P-O bonds and 4 x P=O bonds = 12 x -376 = -4512 plus 4 x 460 = -1840 the total = -6352 kJ
STEP 5: You now have P4O10(g) which must be converted to P4O10(s) - this is the reverse of the vaporisation enthalpy = -93.52 kJ
STEP 6: OK, so you add up all of the steps:
51.9 + 1206 + 2475 - 6352 - 93.52 = -2713 kJ
Last edited by charco; 1 year ago
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#7
(Original post by charco)
STEP 1: You start off with P4(s) molecules which you must convert to P4(g) - this is the enthalpy of vaporisation of phosphorus = +51.9 kJ
STEP 2: You then have to convert P4(g) into 4P(g) - this is six times the bond enthalpy of phosphorus (look at the diagram of the molecule) = 6 x 201 = +1206 kJ
STEP 3: Now you need to convert 5O2(g) into 10O(g) - this is five times the bond enthalpy of oxygen = 5 x 495 kJ = +2475 kJ
STEP 1: You start off with P4(s) molecules which you must convert to P4(g) - this is the enthalpy of vaporisation of phosphorus = +51.9 kJ
STEP 2: You then have to convert P4(g) into 4P(g) - this is six times the bond enthalpy of phosphorus (look at the diagram of the molecule) = 6 x 201 = +1206 kJ
STEP 3: Now you need to convert 5O2(g) into 10O(g) - this is five times the bond enthalpy of oxygen = 5 x 495 kJ = +2475 kJ
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