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vampirchik [111]
2 years ago
11

If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final

Chemistry
1 answer:
mojhsa [17]2 years ago
3 0

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l)  +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

\tt \dfrac{8.5}{78.11}=0.109

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol

Heat transferred to water :

Q=m.c.ΔT

\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC

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Use the given data at 500 K to calculate ΔG°for the reaction
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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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3 years ago
Please help me would really appreciate it thank you :)
tatyana61 [14]

Answer:

e. 8.04*10^4

Explanation:

80.4 g converted to mg is 80,400. 8.04*10^4= 80,400

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