How much heat will be released when 1.48 g of chlorine reacts with excess phosphorus according to the following equation? 2P + 5Cl2 → 2PCl5 ΔH = -886 kJkJ (Your answer should have 3 sig figs.)

-3.70kJ

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,

Δ

H

rxn

, for that particular reaction.

In your case, you have

2

P

(

s

)

+

5

Cl

2

(

g

)

→

2

PCl

5

(

g

)

Δ

H

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

g

⋅

1 mole Cl

2

70.906

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

moles Cl

2

⋅

2 moles PCl

5

5

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

0.008349

moles PCl

5

⋅

886 kJ

2

moles PCl

5

=

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,

Δ

H

rxn

, is equal to

−

3.70 kJ

when

0.008349

moles of product are formed

are equivalent!

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