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faltersainse [42]
3 years ago
11

Calculate the enthalpy change, ΔH, for the process in which 30.1 g of water is converted from liquid at 10.1 ∘C to vapor at 25.0

∘C . For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and Cs = 4.18 J/(g⋅∘C) for H2O(l).
Chemistry
1 answer:
gayaneshka [121]3 years ago
7 0
You can split the process in two parts:

1) heating the liquid water from 10.1 °C to 25.0 °C , and

2) vaporization of liquid water at constant temperature of 25.0 °C.


For the first part, you use the formula ΔH = m*Cs*ΔT

ΔH = 30.1g * 4.18 j/(g°C)*(25.0°C - 10.1°C) = 1,874 J

For the second part, you use the formula ΔH = n*ΔHvap

Where n is the number of moles, which is calculated using the mass and the molar mass of the water:

n = mass / [molar mass] = 30.1 g / 18.0 g/mol = 1.67 mol

=> ΔH = 1.67 mol * 44,000 J / mol = 73,480 J

3) The enthalpy change of the process is the sum of both changes:

ΔH total =  1,874 J + 73,480 J = 75,354 J

Answer: 75,354 J
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The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂ 
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂ 
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H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
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