1.Using the enthalpies of formation given below, calculate H° rxn in kJ, for the

Question

1.Using the enthalpies of formation given below, calculate H°rxn in kJ, for the following reaction.

Report your answer to two decimal places in standard notation.



H2S(g) + 2O2(g) SO3(g) + H2O(l)



H2S (g): -20.60 kJ/mol

O2 (g): 0.00 kJ/mol

SO3 (g): -395.77 kJ/mol

H2O (l): -285.83 kJ/mol

2. Calculate the amount of heat absorbed/released (in kJ) when 22.54 grams of SO3 are produced via the above reaction.

Report your answer to two decimal places, and use appropriate signs to indicate heat flow direction.

Explanation / Answer

1) Given reaction is H2S(g) + 2O2(g) SO3(g) + H2O(l)

H°rxn = Hfo(products) - Hfo( reactants)

= Hfo [SO3(g)] + Hfo [H2O(l)] - { Hfo [H2S(g)]+ 2Hfo [O2(g)]}

=  -395.77 kJ/mol + (-285.83 kJ/mol) - {-20.60 kJ/mol + 2 x0.00 kJ/mol }

= -661 kJ/mol

Therefore, H°rxn = -661 kJ/mol

2)

Given reaction is H2S(g) + 2O2(g) SO3(g) + H2O(l)   H°rxn = -661 kJ/mol   

H°rxn = -ve, it means heat energy is released.

H2S(g) + 2O2(g) SO3(g) + H2O(l)   H°rxn = -661 kJ/mol     

80 g 661 kJ energy is released.

  22.54 g ?

? = (22.54 g / 80 g ) x 661 kJ energy is released

= 186.24 kJ energy is released

Therefore,

186.24 kJ heat energy is released when 22.54 grams of SO3 are produced .

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