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grigory [225]
2 years ago
8

Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2

kJ. What is the heat of formation of SO2 in kilojoules?S (s)+O2 (g)→SO2 (g)?
Chemistry
1 answer:
abruzzese [7]2 years ago
4 0

Answer:

The heat of formation of SO2 is -296.1 kJ

Explanation:

<u>Step 1:</u> Data given

2S (s)+3O2 (g)→2SO3 (g)     ΔH = -790.4 kJ  

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

<u>Step 2</u>: Calculate the heat of formation of SO2

2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ  

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

----------------------------------------------------------------

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

-------------------------------------------------------------------

S (s)+O2 (g)→SO2 (g)

ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ

ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ

The heat of formation of SO2 is -296.1 kJ

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Assume that the amount needed from the 5% acid is x and that the amount needed from the 6.5% acid is y.

We are given that:
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This can be rewritten as:
x = 200 - y .......> equation I

We are also given that:
The concentration of the final solution is 6%
This means that:
5%x + 6.5%y = 6% (x+y)
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0.05 x + 0.065 y = 0.06 (x+y) ............> equation II

Substitute with equation I in equation II and solve for y as follows:
0.05 x + 0.065 y = 0.06 (x+y)
0.05 (200-y) + 0.065 y = 0.06 (200-y+y)
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A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2
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Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

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b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

C_3H_7+5O_2\rightarrow 3CO_2+4H_2O

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

\frac{3}{1}\times 3.85 mol=11.55 mol of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

\frac{4}{1}\times 3.85 mol=15.4 mol of water .

15.4 moles of water are formed from 3.85 mole of propane.

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According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

\frac{4}{5}\times 0.647 mol=0.5176 mol of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

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b. Increasing [C] and [D]:

For a reversible reaction, the value of free energy can be calculated by:

ΔG = ΔG° + RT*lnK

Where ΔG° is the standard value for free energy, R is the gas constant, T is the temperature, and K is the constant of equilibrium, which in this case:

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Answer:

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c 1 V 1 = c 2 V 2

Initial Volume, V1 = 200 mL

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Final Concentration, c2= ?

Solving for c2;

c2 = c1v1 / v2

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c2 = 0.20M

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