Calculate the standard enthalpy change in kJ for the production of 12.2 g of H2O(l) from a reaction with the reactants: H2S(g), O2(g) and the products: H2O(l), SO2(g)?

Question

Jeremiah Adams · Accepted Answer

#DeltaH_"rxn" = -"381 kJ"#

The first thing to do here is write a balanced chemical equation for this reaction

#2"H"_2"S"_text((g]) + 3"O"_text(2(g]) -> 2"SO"_text(2(g]) + 2"H"_2"O"_text((l])#

SInce the standard enthalpy change of reaction, #DeltaH_"rxn"^@#, for this reaction was not provided, you're going to have to calculate it using standard enthalpies of formation

https://tutor.hix.ai

In your case, you have

#"For H"_2"S: " DeltaH_text(f)^@ = -"20.63 kJ/mol"#
#"For O"_2 : " "DeltaH_text(f)^@ = "0 kJ/mol"#
#"For SO"_2: " "DeltaH_text(f)^@ = - "296.84 kJ/mol"#
#"For H"_2"O"_text((l]): DeltaH_text(f)^@ = -"285.8 kJ/mol"#

The standard enthalpy change of reaction can be calculated by using the equation

#color(blue)(DeltaH_"rxn"^@ = sum(n xx Delta_"f products") - sum(m xx DeltaH_"reactants"))" "# , where
#n#, #m# - the number of moles of each product and reactant, respectively

In your case, you would have

#DeltaH_"rxn"^@ = [2color(red)(cancel(color(black)("moles"))) xx (-296.84"kJ"/color(red)(cancel(color(black)("mole")))) + 2color(red)(cancel(color(black)("moles"))) xx (-285.8"kJ"/color(red)(cancel(color(black)("mole"))))] - [2color(red)(cancel(color(black)("moles"))) xx (-20.63"kJ"/color(red)(cancel(color(black)("mole")))) + "3 moles" * 0]#
#DeltaH_"rxn"^@ = -"1165.28 kJ" - (-"41.26 kJ")#
#DeltaH_"rxn" = -"1124.02 kJ"#

Now, this is the enthalpy change of reaction when 2 moles of hydrogen sulfide react with 3 moles of oxygen gas to produce 2 moles of sulfur dioxide and 2 moles of water.

In your case, the mass of water produced by the reaction will help you determine how many moles of each species were actually involved in the reaction.

Use water's molar mass to get

#12.2color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.6772 moles H"_2"O"#

If the reaction will release #"1124.02 kJ"# of heat when 2 moles of water are produced, it follows that you will get

#0.6772color(red)(cancel(color(black)("moles H"_2"O"))) * (-"1124.02 kJ")/(2color(red)(cancel(color(black)("moles H"_2"O")))) = color(green)(-"381 kJ")#

Axel Alvarado · Answer

undefined To calculate the standard enthalpy change for the given reaction, we need to use Hess's law and the enthalpies of formation ($ \Delta H_f^\circ $) for the reactants and products. The balanced chemical equation for the reaction is:

$$ 2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g) $$

First, determine the $ \Delta H_f^\circ $ for each substance involved in the reaction:

$$ \Delta H_f^\circ \text{ for } H_2S(g) = -20.6 \, \text{kJ/mol} $$
$$ \Delta H_f^\circ \text{ for } O_2(g) = 0 \, \text{kJ/mol} $$
$$ \Delta H_f^\circ \text{ for } H_2O(l) = -285.8 \, \text{kJ/mol} $$
$$ \Delta H_f^\circ \text{ for } SO_2(g) = -296.8 \, \text{kJ/mol} $$

Now, calculate the $ \Delta H $ for the reaction using the $ \Delta H_f^\circ $ values:

$$ \Delta H = \sum \Delta H_f^\circ \text{ (products) } - \sum \Delta H_f^\circ \text{ (reactants) } $$

$$ \Delta H = [2 \times (-285.8 \, \text{kJ/mol}) + 2 \times (-296.8 \, \text{kJ/mol})] - [2 \times (-20.6 \, \text{kJ/mol}) + 3 \times (0 \, \text{kJ/mol})] $$

$$ \Delta H = (-571.6 \, \text{kJ/mol} - 593.6 \, \text{kJ/mol}) - (-41.2 \, \text{kJ/mol}) $$

$$ \Delta H = -1165.2 \, \text{kJ/mol} + 41.2 \, \text{kJ/mol} $$

$$ \Delta H = -1124 \, \text{kJ/mol} $$

Now, calculate the $ \Delta H $ for the given amount of $ H_2O(l) $:

$$ \Delta H = -1124 \, \text{kJ/mol} \times \frac{12.2 \, \text{g}}{18.015 \, \text{g/mol}} $$

$$ \Delta H \approx -765.93 \, \text{kJ} $$

So, the standard enthalpy change for the production of $ 12.2 \, \text{g} $ of $ H_2O(l) $ is approximately $ -765.93 \, \text{kJ} $.