# What is the enthalpy change of reaction for this reaction?

##
#2"A" + B rightleftharpoons 2"C" + 2"D"#

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that we can use **Hess' Law** to calculate the standard enthalpy change of reaction,

**Hess' Law** states that the enthalpy change of a reaction is **independent** of the path taken by the reaction.

In your case, you have

#color(blue)(2)"A" + "B" rightleftharpoons color(purple)(2)"C" + color(brown)(2)"D"#

Now, the standard enthalpy change of reaction can be calculated by taking the **sum** of the enthalpy changes of formation of the products multiplied by their respective stoichiometric coefficients and the **sum** of the enthalpy changes of formation of the reactants multiplied by their respective stoichiometric coefficients.

#color(blue)(ul(color(black)(DeltaH_"rxn"^@ = sum_i n xx DeltaH_"f i"^@ - sum_j m xx DeltaH_"f j"^@)))#

Here

#n# is the stoichiometric coefficient of a product#m# is the stoichiometric coefficient of a reactant

So, you know that you have

#color(blue)(2)# molesof#"A"# and#1# moleof#"B"# on the reactants' side#color(purple)(2)# molesof#"C"# and#color(brown)(2)# molesof#"C"# on the products' side

You can say that for the products, you have

#sum_j m xx DeltaH_"f j"^@ = color(purple)(2)color(red)(cancel(color(black)("moles"))) * (+215 "kJ"/color(red)(cancel(color(black)("mol")))) + color(brown)(2)color(red)(cancel(color(black)("moles"))) * (-523 "kJ"/color(red)(cancel(color(black)("mol"))))#

#= "430 kJ" + (-"1046 kJ")#

# = -"616 kJ"#

Similarly, you can say that for the reactants', you have

#sum_i n xx DeltaH_"f i"^@ = color(blue)(2)color(red)(cancel(color(black)("moles"))) * (-225 "kJ"/color(red)(cancel(color(black)("mol")))) + 1color(red)(cancel(color(black)("mole"))) * (-403 "kJ"/color(red)(cancel(color(black)("mol"))))#

# = -"450 kJ" + (-"403 kJ")#

# = -"853 kJ"#

Therefore, you can say that the standard enthalpy change of reaction is equal to

#DeltaH_"rxn"^@ = -"853 kJ" - (-"616 kJ")#

#color(darkgreen)(ul(color(black)(DeltaH_"rxn"^@ = -"237 kJ")))#

The answer is rounded to three **sig figs**.