Question

Question #69013

Answer 1

`"3.70 kJ"`

Explanation:

Notice that the problem provides you with the thermochemical equation for this reaction.

A thermochemical equation is simply a balanced chemical equation that includes the enthalpy change of reaction, `DeltaH_"rxn"`, for that particular reaction.

In your case, you have

`2"P"_ ((s)) + 5"Cl"_ (2(g)) -> 2"PCl"_ (5(g))" "DeltaH_text(rxn) = -"886 kJ"`

This tells you that when the reaction produces two moles of phosphorus pentachloride, `"PCl"_5`, a total of `"886 kJ"` of heat are being given off `->` the reaction is exothermic.

Now, you need to figure out how much heat will be given off when `"1.48 g"` of chlorine gas reacts with excess phosphorus.

The fact that phosphorus is in excess tells you that chlorine will act as a limiting reagent, i.e. it will be completely consumed by the reaction.

Your goal now is to determine how many moles of chlorine gas you have in that sample. To do that, use its molar mass

`1.48 color(red)(cancel(color(black)("g"))) * "1 mole Cl"_2/(70.906color(red)(cancel(color(black)("g")))) = "0.02087 moles Cl"_2`

Use the `5:2` mole ratio that exists between chlorine gas and phosphorus pentachloride to determine how many moles of the latter will be produced when `0.02807` moles of chlorine take part in the reaction.

`0.02807 color(red)(cancel(color(black)("moles Cl"_2))) * ("2 moles PCl"_5)/(5color(red)(cancel(color(black)("moles Cl"_2)))) = "0.008349 moles PCl"_5`

Now, you know that the reaction gives off `"886 kJ"` of heat when `2` moles of phosphorus pentachloride are formed. This means that when `0.008349` moles of the product are formed, the reaction will give off

`0.008349color(red)(cancel(color(black)("moles PCl"_5))) * "886 kJ"/(2color(red)(cancel(color(black)("moles PCl"_5)))) = color(green)("3.70 kJ")`

The answer is rounded to three sig figs.

Remember, these two statements

The reaction gives off `"3.70 kJ"` of heat when `0.008349` moles of product are formed

and

The enthalpy change of reaction, `DeltaH_"rxn"`, is equal to `-"3.70 kJ"` when `0.008349` moles of product are formed

are equivalent!