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Over [174]
3 years ago
8

Predict the sign and calculate ΔS° for a reaction. Close Problem Consider the reaction H2CO(g) + O2(g)CO2(g) + H2O(l) Based upon

the stoichiometry of the reaction the sign of Sºrxn should be _________ . Using standard thermodynamic data (in the Chemistry References), calculate Sºrxn at 25°C. Sºrxn = J/K•mol
Chemistry
1 answer:
zhenek [66]3 years ago
7 0

Answer:

\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}

Based upon the stoichiometry of the reaction the sign of Sºrxn should be <u>negative</u>

Explanation:

Consider the reaction:

H2CO(g) + O2(g) --------> CO2(g) + H2O(l)

Using standard thermodynamic data;

Based upon the stoichiometry of the reaction the sign of Sºrxn should be _________ . calculate Sºrxn at 25°C. Sºrxn = J/K•mol

At standard thermodynamic data

\mathtt{S^0_{rxn} = \sum S^0 _{product} - \sum S^0 _{reactant}}

S^0(CO_2) = 213.79 J/mol.K

S^0(H_2O)=  69.95  J/mol.K

S^0 ({H_2CO}) = 218.95 J/mol.K

S^0 (O_2)  = 205.2 J/mol.K

\mathtt{S^0_{rxn} = (213.79 + 69.95) J/mol.K  - (218.95+ 205.2) J/mol.K}

\mathtt{S^0_{rxn} = (283.74) J/mol.K  - (424.15) J/mol.K}

\mathbf{S^0_{rxn} = -140.41 \ J/mol.K}

Based upon the stoichiometry of the reaction the sign of Sºrxn should be <u>negative</u>

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The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
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Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

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The half oxidation-reduction reaction will be :

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Now we have to calculate the Gibbs free energy.

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n = number of electrons to balance the reaction = 2

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Now put all the given values in this formula, we get the Gibbs free energy.

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Read 2 more answers
A sample of propane(c3h8)has 3.84x10^24 H atoms.
deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}

<u>= 4.75 × 10²³ molecules of Propane</u>.

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

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

<u>= 14.25 × 10²³ </u>

<u>________________________________________</u>

<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

\mathsf{ \frac{44 }{6.02 \times  {10}^{23} } \times 4.75 \times  {10}^{23}  }

\mathsf{  = \frac{44 }{6.02 \times   \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}}  }

\mathsf{  = \frac{44 }{6.02 } \times 4.75   }

<u>= 34.72 g</u>

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