Question

Using the bond energies below, calculate an estimate of AHrxn for the gas phase reaction: QX3 + 3H20 => Q(OH)3 + 3HX Do not enter units with your answer. Bond BE (kJ/mol) Q-X 240 O-H 464 Q-O 359 H-X 449

Answer 1

To calculate the approximate enthalpy change for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The approximate enthalpy change is -3818 kJ.

Step-by-step explanation:

To calculate the approximate enthalpy change (ΔH) for the given reaction, we need to consider the bond energies of the bonds in the reactants and products. The reaction is:

QX3 + 3H20 => Q(OH)3 + 3HX

We can calculate the energy absorbed or released by breaking and forming these bonds.

First, we calculate the energy required to break the bonds in the reactants:

• For Q-X bonds: 3 mol x 240 kJ/mol = -720 kJ
• For O-H bonds: 3 mol x 464 kJ/mol = -1392 kJ

Then, we calculate the energy released when the bonds in the products are formed:

• For Q-O bonds: 1 mol x 359 kJ/mol = -359 kJ
• For H-X bonds: 3 mol x 449 kJ/mol = -1347 kJ

Finally, we sum up the energy changes:

• -720 kJ + (-1392 kJ) + (-359 kJ) + (-1347 kJ) = -3818 kJ

Therefore, the approximate enthalpy change for the reaction is -3818 kJ.

Answer 2

The reaction enthalpy is 1080 kJ/mol

Step-by-step explanation:

The given reaction is:

`QX_(3) + 3H_(2)0 \rightarrow Q(OH)_(3) + 3HX`

The reaction enthalpy is given by the difference between the number of bonds broken and the number of bonds formed.

`\Delta H_(rxn)=\sum (bonds\ broken)-\sum (bonds\ formed)`

For the given reaction:

`\Delta H_(rxn)=[3(Q-X)+6(O-H)]-[3(Q-O)+3(H-X)]`

from the given bond energies:

`\Delta H_(rxn)=[3(240)+6(464)]-[3(359)+3(449)]=1080kJ/mol`