Question

The thermochemical equation for the chemical reaction between phosphorus and bromine is given below. How much heat is produced when $ 2.63 $ g of phosphorus is allowed to react with 40 g of bromine?
 $ {{\text{P}}_{\text{4}}}\left( {\text{s}} \right){\text{ + 6B}}{{\text{r}}_{\text{2}}}\left( {\text{l}} \right) \to 4{\text{PB}}{{\text{r}}_{\text{3}}}\left( {\text{g}} \right),{{\Delta {\rm H} = - 486Kj}} $ .

Answer 1

Hint: The heat of reaction is defined as the amount of energy that is released or required when the reactants are transformed into products. It is expressed in the units of KiloJoules and the molar enthalpy in the units of $ {\text{Kilojoules/ mol}} $ . We shall use the stoichiometric coefficients to find the number of moles reacting according to the given weight and use it to find the heat.

Complete step by step Solution
According to the question, the equation for the chemical reaction between phosphorus and bromine is:
 $ {{\text{P}}_{\text{4}}}\left( {\text{s}} \right){\text{ + 6B}}{{\text{r}}_{\text{2}}}\left( {\text{l}} \right) \to 4{\text{PB}}{{\text{r}}_{\text{3}}}\left( {\text{g}} \right),{{\Delta {\rm H} = }} - {\text{486Kj}} $
As per the reaction, 1 mole of phosphorus reacts with 6 moles of bromine to form 4 moles phosphorous tetrabromide. The molecular weight of phosphorus = $ \left( {4 \times 31} \right) = 124 $ $ {\text{g/mol}} $ .
Molecular weight of bromine = $ \left( {80 \times 2} \right) = 160 $ $ {\text{g/mol}} $
Molecular weight of phosphorus tribromide = $ \left[ {\left( {80 \times 4} \right) + 31} \right] = 351 $ $ {\text{g/mol}} $
As per the definition of mole, one mole of any substance is equal to the molecular weight of that substance in grams. Therefore,
1 mole of phosphorus = 124 grams
6 moles of bromine = $ \left( {6 \times 160} \right) = 960 $ grams
And 4 moles of diphosphorus tetrabromide = $ \left( {4 \times 351} \right) = 1404 $ grams
Now, 124 grams of phosphorus requires 960 grams of bromine.
Therefore, $ 2.63 $ g of phosphorus would require $ \dfrac{{960}}{{124}} \times 2.63 = 20.361 $ grams of bromine.
But there are 40 grams of bromine in the reaction medium.
Therefore, the limiting reagent in the medium is Phosphorous as it is completely consumed in the reaction.
As the reaction results in the release of $ - 486 $ Kilojoules of heat for 124 grams of phosphorus, the heat released for the reaction of $ 2.63 $ g of phosphorus, is equal to
 $ \dfrac{{ - 486}}{{124}} \times 2.63 = - 10.30 $ KiloJoules of heat.
So the heat produced when $ 2.63 $ g of phosphorus is allowed to react with 40 g of bromine is $ - 10.30 $ KiloJoules.

Note
If the heat of reaction is negative, then it indicates an exothermic reaction change while the heat of reaction being positive indicates that the reaction is endothermic in nature and would require energy from outside.