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scoundrel [369]
3 years ago
11

How many moles of hydrogen will be produced in this reaction when 34.5 moles of hydrogen iodide are decomposed?

Chemistry
1 answer:
irina1246 [14]3 years ago
6 0
<h3><u>Answer</u>;</h3>

17.25 moles of hydrogen

<h3><u>Explanation;</u></h3>

The equation for the reaction;

2HI (g) → H₂(g) + i₂ (g)

34.5 moles of Hydrogen iodide are decomposed;

The mole ratio of HI to Hydrogen = 2 : 1

Therefore;

moles of hydrogen = (34.5/2) ×1

                                = 17.25 moles

Therefore; 17.25 moles of hydrogen will be produced when 34.5 moles of Hydrogen iodide are decomposed

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

Explanation :

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\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

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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)]

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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)]

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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

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

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