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Chemistry questions

Hey guys i need help with this question any help will be much appreciated

(a) Two forms of sulfur are rhombic and monoclinic.
enthalpies of combustion to SO2 are
AH°(rhombic) = -298.5 kJ mol -~
AH°(monoclinic) = -301.2 kJ mol -I
At 298 K the standard
Estimate the standard enthalpy change for the process
S(rhombic) ~ S(monoclinic). [5]

(b) Arrange the following elements in order of increasing molar entropy, assuming
standard states at room temperature and pressure: lead, neon, silicon, bromine,
oxygen. Briefly state the reasons for your choice. [5]

c) What is the minimum wavelength of electromagnetic radiation required to cleave
the H-H bond in the H2 molecule, which has a bond energy of 420 kJ mol-l? [5]

(d) Explain the following observations. H2 has an H-H distance of 0.74 /~ and a
bond energy of 420 kJ mol -l whilst H~ has an H-H distance of 1.06 A and a
bond energy of 240 kJ mol-k [5]
For part C, i would reckon that you use E=hf (energy = planck's constant x frequency), then use v=λf to find the wavelength of the radiation (v = 3x10^8 ms-1)
Reply 2
A)


[1] S(rhombic) +O2 → SO2 ΔH° = -298.5 kJ mol-1
[2] S(monocyclic) + O2 → SO2 ΔH° = -301.2 kJ mol-1

S(rhombic) → S(monocyclic) ΔH° = ?

For this one, the most straight forward thing to do is to "substract" the second equation from the first one, making the reactants, products, and the products, reactants and changing the ΔH° sign from negative to positive.

B)

Neon should be the one with highest molar entropy, because i'ts a monoatomic gas. Oxygen would have the second highest, due to it being a diatomic gas. Bromine would be next, being a diatomic liquid. I'm no totally sure about the order for lead/silicon out of my mind, but I think I would put lead as the one with least entropy due to it being a metal, whether silicon is a metalloid.

C)

You need to first get the energy per bond first, then do what pigetrational said. Also, it should'nt be "minimum wavelenght", but "minimum frequency" or "maximum wavelenght", as with higher wavelenght you have less energy/photon.

D)

I'm not sure what that question is about, can you please rephrase it?
Qwaps


B)

Neon should be the one with highest molar entropy, because i'ts a monoatomic gas. Oxygen would have the second highest, due to it being a diatomic gas. Bromine would be next, being a diatomic liquid. I'm no totally sure about the order for lead/silicon out of my mind, but I think I would put lead as the one with least entropy due to it being a metal, whether silicon is a metalloid.



entropy increases with increased mass and molecular complexity. Oxygen would have greater molar entropy than neon. Silicon is a giant molecular lattice solid and has very low entropy.

oxygen > neon > bromine > lead > silicon




Qwaps

I'm not sure what that question is about, can you please rephrase it?


me neither! What is H~ ??
Reply 4
Sorry guys for question (d) Explain the following observations. H2 has an H---H distance of 0.74 A (this is the A with a circle on top meaning atomic seaparation a really small ditance measurement) and a bond energy of 420 kJ /mol whilst H+ has an H----H distance of 1.06 A and a bond energy of 240 kJ /mol [5]

Oh btw for c) i got 4.733 X10^-31 for the wavelength....is tht correct?
and for a) i got 11.7 kJ/mol

Thnks for your help so far it has been really useful :smile:
Latkan
Sorry guys for question (d) Explain the following observations. H2 has an H---H distance of 0.74 A (this is the A with a circle on top meaning atomic seaparation a really small ditance measurement) and a bond energy of 420 kJ /mol whilst H+ has an H----H distance of 1.06 A and a bond energy of 240 kJ /mol [5]

Oh btw for c) i got 4.733 X10^-31 for the wavelength....is tht correct?
and for a) i got 11.7 kJ/mol

Thnks for your help so far it has been really useful :smile:


This is meaningless. One particle cannot have either a distance between itself nor a bond strength. Are you sure you dont mean H2+?
For part C I get this:

420×10006.023×1023=6.97×1019J\frac {420 \times 1000}{6.023 \times 10^{23}} = 6.97 \times 10^{-19} \mathrm{J} = Energy to break one H2 atom into 2H·

Then use E=hfE = hf, so f=6.97×10196.63×1034=1.05×1015Hzf = \frac {6.97 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.05 \times 10^{15} \mathrm{Hz}

Then from v=λfv={\lambda}{f},
λ=3×1081.05×1015=285×109m\lambda = \frac{3 \times 10^8}{1.05 \times 10^{15}} = 285 \times 10^{-9} \mathrm{m}, which is in the range of UV light.

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