as level chemistry help 😢😢😢
(b) Some bond energies are given.
bond bond energy / kJmol–1
S=O (in SO2) 534
O=O 496
Use the data, and the enthalpy change for the conversion of sulfur dioxide into sulfur trioxide,
to calculate a value for the S=O bond energy in SO3.
S=O bond energy in SO3 = .............................. kJmol–1 [2]
bond bond energy / kJmol–1
S=O (in SO2) 534
O=O 496
Use the data, and the enthalpy change for the conversion of sulfur dioxide into sulfur trioxide,
to calculate a value for the S=O bond energy in SO3.
S=O bond energy in SO3 = .............................. kJmol–1 [2]
Start by writing out the balanced chemical equation, then work out how much energy is required to break all the bonds, then how much energy is released when all the new bonds are formed 
the equation is balanced
the ans that i have calculated is different form the original ans.
Original post by shah kaivan
the ans that i have calculated is different form the original ans.
You really should post the whole question
Did they give you the balanced equation? What value have they given the enthalpy change of formation of SO3.What answer did you get?
(edited 4 years ago)
ok
s=o 1068 + o=o 496 == s=o 1602
1068+496-1602
s=o 1068 + o=o 496 == s=o 1602
1068+496-1602
so my ans is -38
Original post by shah kaivan
so my ans is -38
Bond enthalpies are always positive so you shouldn't need the answer to tell you something has gone wrong.
Step 1: Balanced chemical equation
SO2 + 0.5O2 → SO3 where ΔHf = -396 kJ mol-1 (value taken from a data book, the value you have might be different)
Step 2: Work out energy needed to break bonds
SO2 contains 2 x (S=O) bond, O2 has 1 (O=O) bond but have only half a mole of this!
Bonds broken = 2 x 534 + 0.5 x 496 = 1316 kJ mol-1
Step 3: Work out energy needed to form the new bonds
ΔHf = bonds broken - bonds formed
-396 kJ mol-1= 1316 kJ mol-1 - 3(S=O)
Can you finish it off?
Seems not
(edited 4 years ago)
this is in the mark scheme
−196 + 6S=O = (4 × 534) + 496 1
S=O = 2828 / 6 = 471(.3)
−196 + 6S=O = (4 × 534) + 496 1
S=O = 2828 / 6 = 471(.3)
how did he multiply it by 6
Original post by shah kaivan
how did he multiply it by 6
It looks like they've used 2SO2 + O2 → 2SO3 as their balanced equation, which is fine but to do that you also need to use 2ΔHf
do you have mark scheme of save my exam website
in this website topical past papers are given but ms is not free
2.00 moles of nitrogen and 3.00 moles of hydrogen were put in a vessel and left to reach
equilibrium.
At equilibrium, the pressure was 2.00 × 107Pa and the mixture contained 1.60 moles of
ammonia.
Calculate Kp. Include the units.
equilibrium.
At equilibrium, the pressure was 2.00 × 107Pa and the mixture contained 1.60 moles of
ammonia.
Calculate Kp. Include the units.
my ans
n2 h2 2nh3
2 3 0
Y X 1.60
how to find x and y
n2 h2 2nh3
2 3 0
Y X 1.60
how to find x and y
Original post by shah kaivan
(b) Some bond energies are given.
bond bond energy / kJmol–1
S=O (in SO2) 534
O=O 496
Use the data, and the enthalpy change for the conversion of sulfur dioxide into sulfur trioxide,
to calculate a value for the S=O bond energy in SO3.
S=O bond energy in SO3 = .............................. kJmol–1 [2]
bond bond energy / kJmol–1
S=O (in SO2) 534
O=O 496
Use the data, and the enthalpy change for the conversion of sulfur dioxide into sulfur trioxide,
to calculate a value for the S=O bond energy in SO3.
S=O bond energy in SO3 = .............................. kJmol–1 [2]
Is this your homework question from school?
Sadly your teacher would be disappointed with you asking strangers to solve the question for you
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