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Ksenya-84 [330]
3 years ago
15

Which one of the following equations is properly balanced? a. CH 3CHO + 3O 2 → 2CO 2 + 2H 2O b. NH 4NO 3 → 2H 2O + N 2 c. Na 2CO

3 + 2H 2SO 4 → Na 2SO 4 + 2H 2O + CO 2 d. 2Na 2SO 4 + 3Bi(NO 3) 3 → Bi 2(SO 4) 3 + 9NaNO 3 e. Sn + 4HNO 3 → SnO 2 + 4NO 2 + 2H 2O
Chemistry
1 answer:
Gennadij [26K]3 years ago
3 0

Answer:

The chemical equation that is properly balanced is;

e. Sn + 4HNO₃ → SnO₂ + 4NO₂ + 2H₂O

Explanation:

To solve the question, we note the reactions thus and we look for a reason why the reaction is not balanced as follows.

a. CH₃CHO + 3O₂ → 2CO₂ + 2H₂O

Here we have 7 oxygen atoms in the reactant and 6 in the products

Not balanced

b. NH₄NO₃ → 2H₂O + N₂

Here we have three oxygen in the reactant and four in the products = Not balanced

c. Na₂CO₃ + 2H₂SO₄ → Na₂SO₄ + 2H₂O + CO₂

Here we have 11 oxygen atoms in the reactant and 8 in the products = Not balanced

d. 2Na₂SO₄ + 3Bi(NO₃)₃ → Bi₂(SO 4)₃ + 9NaNO₃

Here  we have 2 Na in the reactants and 9 Na in the products = Not balanced

e. Sn + 4HNO₃ → SnO₂ + 4NO₂ + 2H₂O

Here we have equal number of products and reactants = Balanced chemcal reaction equation.

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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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