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Calculate ΔH o rxn for the following: a. SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(l) ΔH o f [SiO2 (s)] = −910.9 kJ/mol ΔH o f [HF (g)] = −273 kJ/mol ΔH o f [SiF4 (g)] = −1,614.9 kJ/mol ΔH o f [H2O (l)] = −285.840 kJ/mol b. C2H6(g) + O2(g) CO2(g) + H2O(g)

Respuesta :

Answer : The value of [tex]\Delta H^o_{rxn}[/tex] for the reaction is -183.68 kJ/mole

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as [tex]\Delta H^o[/tex]

The equation used to calculate enthalpy change is of a reaction is:  

[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]

The equilibrium reaction follows:

[tex]SiO_2(s)+4HF(g)\rightleftharpoons SiF_4(g)+2H_2O(l)[/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta H^o_{rxn}=[(n_{(SiF_4)}\times \Delta H^o_f_{(SiF_4)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(SiO_2)}\times \Delta H^o_f_{(SiO_2)})+(n_{(HF)}\times \Delta H^o_f_{(HF)})][/tex]

We are given:

[tex]\Delta H^o_f_{(SiO_2(s))}=-910.9kJ/mol\\\Delta H^o_f_{(HF(g))}=-273kJ/mol\\\Delta H^o_f_{(SiF_4(g))}=-1614.9kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.840kJ/mol[/tex]

Putting values in above equation, we get:

[tex]\Delta H^o_{rxn}=[(1\times -1614.9)+(2\times -285.840)]-[(1\times -910.9)+(4\times -273)]=-183.68kJ/mol[/tex]

Therefore, the value of [tex]\Delta H^o_{rxn}[/tex] for the reaction is -183.68 kJ/mole

pstnonsonjoku

The heat of reaction is -183.78 kJ/mol.

We have the equation of the reaction as follows;

SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

The following information are available;

  • ΔH o f [SiO2 (s)] = −910.9 kJ/mol
  • ΔH o f [HF (g)] = −273 kJ/mol
  • ΔH o f [SiF4 (g)] = −1,614.9 kJ/mol
  • ΔH o f [H2O (l)] = −285.840 kJ/mol

Since;

ΔH o rxn = ∑ΔH o f(products) - ΔH o f(reactants)

ΔH o rxn = ∑[(1 × (−1,614.9 kJ/mol)) + (2 × (−285.840 kJ/mol))] - ∑[(1 × ( −910.9 kJ/mol )) + (4 × (−273 kJ/mol))

ΔH o rxn = (-2186.68 kJ/mol) - (-2002.9 kJ/mol)

ΔH o rxn = -183.78 kJ/mol

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