Question

How to calculate heat release given moles of precipitate and delta H of the reaction?

Ex: if the Delta H reaction =-55.5kJ, how much heat is released when 3.31 moles of precipitate are formed?

Answer 1

Ans. Note that dH of the reaction = -55.5 kJ but NOT “-55.5 kJ/mol”. The value of MOLAR enthalpy, dH_molar may vary depending on the given reaction. So, follows the steps explained below-

# Note the following reactions-

#1. A(aq) + B(aq) ------> C(s)         ; dH = -55.5 kJ

The dH for the reaction, i.e. formation of 1 mol precipitate, C(s), is -55.5 kJ.

So,

dH_molar = dH / moles of precipitate formed = (-55.5 kJ) / 1 mol = -55.5 kJ mol^-1

Now,

Theoretical amount heat released during reaction = dH_molar x Moles of ppt. formed

                                                                        = (-55.5 kJ mol^-1) x 3.31 mol

                                                                        = -183.705 kJ

#2. A(aq) + B(aq) ------> 2C(s)      ; dH = -55.5 kJ

The dH for the reaction, i.e. formation of 2 mol precipitate, C(s), is -55.5 kJ.

So,

dH_molar = dH / moles of precipitate formed = (-55.5 kJ) / 2 mol = -27.75 kJ mol^-1

Now,

Theoretical amount heat released during reaction = dH_molar x Moles of ppt. formed

                                                                        = (-27.75 kJ mol^-1) x 3.31 mol

                                                                        = -91.8525 kJ

# Conclusion: If the given dH is in terms of “-55.5 kJ / mol” follow #1.

However, if dH of the reaction is given only as “-55.5 kJ”, first calculate dH_molar, then calculate the expected/theoretical amount of heat released.