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-BARSIC- [3]
3 years ago
14

Sulfur dioxide, SO 2 ( g ) , can react with oxygen to produce sulfur trioxide, SO 3 ( g ) , by the reaction 2 SO 2 ( g ) + O 2 (

g ) ⟶ 2 SO 3 ( g ) The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol Calculate the amount of energy in the form of heat that is produced when a volume of 2.67 L of SO 2 ( g ) is converted to 2.67 L of SO 3 ( g ) according to this process at a constant pressure and temperature of 1.00 bar and 25.0 °C . Assume ideal gas behavior.
Chemistry
1 answer:
Alex3 years ago
5 0

Answer:

ΔHorxn = - 11.79 KJ

Explanation:

2 SO 2 ( g ) + O 2 ( g ) ⟶ 2 SO 3 ( g )

The standard enthalpies of formation for SO 2 ( g ) and SO 3 ( g ) are Δ H ∘ f [ SO 2 ( g ) ] = − 296.8 kJ / mol Δ H ∘ f [ SO 3 ( g ) ] = − 395.7 kJ / mol

From the reaction above, 2 mol of SO2 reacts to produce 2 mol of SO3. Assuming ideal gas behaviour,

1 mol = 22.4l

x mol = 2.67l

Upon cross multiplication and solving for x;

x = 2.67 / 22.4 = 0.1192 mol

0.1192 mol of SO2 would react to produce 0.1192 mol of SO3.

Amount of heat is given as;

ΔHorxn = ∑mΔHof(products) − ∑nΔHof(reactants)

Because O2(g) is a pure element in its standard state, ΔHοf [O2(g)] = 0 kJ/mol.

ΔHorxn = 0.1192 mol * (− 395.7 kJ / mol) - 0.1192 mol * ( − 296.8 kJ / mol)

ΔHorxn = - 47.17kj + 35.38kj

ΔHorxn = - 11.79 KJ

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The vapor pressure of water at 25.0°c is 23.8 torr. determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.
svp [43]
Q: A
according to this formula, we can get the mole fraction of water (n):
P(solu) = n Pv(water)
when we have Pv(solu) = 22.8 and Pv(water) = 23.8 so by substitution:
22.8 = n * 23.8
n= 0.958
- we need to get the moles of glucose:
moles of water = 500 g(mass weight) / 18 (molar weight)= 27.7 mol
n = moles of water / ( moles of water + moles of glucose)
0.958   = 27.7 / ( 27.7+ moles of glucose)
0.958 moles of glucose + 26.5 = 27.7
0.968 moles of glucose = 1.2
moles of glucose = 1.253 mol
∴ the mass of glucose = no.of glucose moles x molar mass 
                                      = 1.253 x 180 = 225.5 g
Q: B
here we also need to get n (mole fraction of water )by using this formula:
Pv(solu) = n Pv(water)
when we have Pv(solu)=132 & Pv(water)=150 so, by substition:
132= n * 150
n = 0.88
so, mole fraction of solution = 1 - 0.88 = 0.12
and we can get after that the moles of water = (mass weight / molar mass)
- no.moles of water = 85 g / 18 g/mol = 4.7 moles
- total moles in solution = moles of water / moles fraction of water 
                                        = 4.7 / 0.88 = 5.34 moles 
∴ moles of the solution = total moles in solu - moles of water 
                                       = 5.34 - 4.7 = 0.64 moles solute
∴ the molar mass of the solute = mass weight of solute / no.of moles of solute
                                                    = 53.8 / 0.64 = 84 g/mole

Q: C

moles of urea (NH2)2 CO = mass weight / molar mass
                                           = 4.49 g / 60 g /mol
                                           = 0.07 mol
moles of methanol = mass weight / molar mass 
                                 = 39.9  g / 32  g/mol = 1.25 mol
moles fraction of methanol = moles of methanol / (moles of methanol + moles of urea )
moles fraction of methanol = 1.25 / ( 1.25+0.07) = 0.95
by substitution in Pv formula we will be able to get the vapour pressure of the solu :
Pv(solu) = n P°v
Pv(solu) = 0.95 * 89 mm Hg 
∴Pv(solu) = 84.55 mmHg


 
7 0
3 years ago
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