Calculate the ƒ¢H�‹ rxn for the following reaction.
ƒ¢H�‹ f [H2S(g)] = -21 kJ/mol; ƒ¢H�‹ f [SO2(g)] = -297 kJ/mol; ƒ¢H�‹ f [H2O(l)] = -285.8 kJ/mol;
2H2S(g) + SO2(g) �¨3S(s) + 2H2O(l)
a. 232.6 kJ
b. 910 kJ
c. -232.6 kJ
d. -910 kJ
To calculate the ∆H°rxn for the given reaction, you can use the equation:
∆H°rxn = ∑∆H°f (products) - ∑∆H°f (reactants)
In this equation, ∑∆H°f (products) represents the sum of the standard enthalpies of formation of the products, and ∑∆H°f (reactants) represents the sum of the standard enthalpies of formation of the reactants.
Given:
∆H°f [H2S(g)] = -21 kJ/mol
∆H°f [SO2(g)] = -297 kJ/mol
∆H°f [H2O(l)] = -285.8 kJ/mol
Now let's calculate the values for the products and reactants:
Products:
3S(s) + 2H2O(l)
Reactants:
2H2S(g) + SO2(g)
Next, calculate the sum of the standard enthalpies of formation for the products:
∑∆H°f (products) = (3 x 0 kJ/mol) + (2 x -285.8 kJ/mol) = -571.6 kJ/mol
Then, calculate the sum of the standard enthalpies of formation for the reactants:
∑∆H°f (reactants) = (2 x -21 kJ/mol) + (-297 kJ/mol) = -339 kJ/mol
Finally, substitute the values into the equation ∆H°rxn = ∑∆H°f (products) - ∑∆H°f (reactants):
∆H°rxn = (-571.6 kJ/mol) - (-339 kJ/mol) = -232.6 kJ/mol
So, the ∆H°rxn for the given reaction is -232.6 kJ/mol. Therefore, the correct answer is option c. -232.6 kJ.