Given the following information: Calculate the change in enthalpy for: 2Li(s)+ 2HBr(g)$\to$H₂(g)+ 2LiBr(s) A)262 kJ B)-602 kJ C)-517 kJ D)-992 kJ E)none of these

Question

Quizplus · Accepted Answer

First, we need to calculate the enthalpy change for the sublimation of 2 moles of Li(s):
ΔH(sublimation) = 2 mol x 166 kJ/mol = 332 kJ/mol
Next, we need to calculate the enthalpy change for the dissociation of 2 moles of HBr(g):
ΔH(dissociation) = 2 mol x 363 kJ/mol = 726 kJ/mol
Then, we need to calculate the enthalpy change for the formation of 2 moles of LiBr(s):
ΔH(formation) = -2 x 809 kJ/mol = -1618 kJ/mol
Finally, we need to calculate the enthalpy change for the formation of 2 moles of H2(g):
ΔH(formation) = 2 x (0 kJ/mol - 432 kJ/mol) = -864 kJ/mol
Now we can add up all the enthalpy changes to obtain the overall enthalpy change:
ΔH = [(2 x 166) + (2 x 363) + (-2 x 809) + (2 x (-432))] kJ/mol
ΔH = -602 kJ/mol
Therefore, the best choice is B, which is the calculated enthalpy change of -602 kJ.

Given the Following Information: Calculate the Change in Enthalpy

Given the Following Information:

Calculate the Change in Enthalpy