Question

1.Using the enthalpies of formation given below, calculate
ΔH°_{rxn} in kJ, for the following reaction.

Report your answer to two decimal places in standard
notation.

H_{2}S(g) + 2O_{2}(g) → SO_{3}(g) +
H_{2}O(l)

H_{2}S (g): -20.60 kJ/mol

O_{2} (g): 0.00 kJ/mol

SO_{3} (g): -395.77 kJ/mol

H_{2}O (l): -285.83 kJ/mol

2. Calculate the amount of heat absorbed/released (in kJ) when
22.54 grams of SO_{3} are produced via the above
reaction.

Report your answer to two decimal places, and use appropriate signs
to indicate heat flow direction.

Answer #1

1) Given reaction is H2S(g) + 2O2(g) → SO3(g) + H2O(l)

ΔH°rxn = **ΔH**_{f}^{o}(products) -
**ΔH**_{f}^{o}( reactants)

= **ΔH**_{f}^{o} [SO3(g)]
+ **ΔH**_{f}^{o} [H_{2}O(l)] -
{ **ΔH**_{f}^{o} [H2S(g)]+
2**ΔH**_{f}^{o} [O2(g)]}

= -395.77 kJ/mol + (-285.83 kJ/mol) - {-20.60 kJ/mol + 2 x0.00 kJ/mol }

= -661 kJ/mol

Therefore, ΔH°rxn = -661 kJ/mol

2)

Given reaction is H2S(g) + 2O2(g) → SO3(g) + H2O(l) ΔH°rxn = -661 kJ/mol

ΔH°rxn = -ve, it means heat energy is released.

H2S(g) + 2O2(g) → SO3(g) + H2O(l) ΔH°rxn = -661 kJ/mol

80 g 661 kJ energy is released.

22.54 g ?

? = (22.54 g / 80 g ) x 661 kJ energy is released

= 186.24 kJ energy is released

Therefore,

186.24 kJ heat energy is released when 22.54 grams of SO3 are produced .

Use standard enthalpies of formation to calculate ΔH∘rxn for the
following reaction: SO2(g)+12O2(g)→SO3(g) ΔH∘rxn =

Use the standard reaction enthalpies given below to determine
ΔH°rxn for the following reaction:
2 S(s) + 3 O2(g) → 2 SO3(g) ΔH°rxn = ?
Given: SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ 2 SO2(g) + O2(g)
→ 2 SO3(g) ΔH°rxn = -197.8 kJ
Please explain in detail.

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equation:
2KI(s) + Br2(l) → 2KBr(s) +
I2(s) ΔHr° =
-131.80 kJ
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2H2S(g) + 3O2(g) → 2SO2(g) +
2H2O(l)
H2S (g): -20.60...

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m =
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Express your answer using five significant figures.
ΔH∘rxn =
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kJ
Compound
ΔH⁰f
(kJ/mol)
TiCl4(g)
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H2O(g)
−241.8
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−944.0
HCl(g)
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Report your answer to two decimal places.
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8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s)
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The standard heat of formation, ΔH∘f, is defined as the
enthalpy change for the formation of one mole of substance from its
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values can be used to calculate the enthalpy change of any
reaction.
Consider, for example, the reaction
2NO(g)+O2(g)⇌2NO2(g)
with heat of formation values given by the following table:
Substance
ΔH∘f
(kJ/mol)
NO(g)
90.2
O2(g)
0
NO2(g)
33.2
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Calculate the standard enthalpy change, ΔH°rxn, in kJ for the
following chemical equation, using only the thermochemical
equations below: 4KO2(s) + 2H2O(l) → 4KOH(aq) + 3O2(g) Report your
answer to three significant figures in scientific notation.
Equations: ΔH°rxn
(kJ)
4K(s) + O2(g) → 2K2O(s)
-726.4
K(s) + O2(g) → KO2(s) -284.5
K2O(s) + H2O(l) → 2KOH(aq) -318

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