Question

Using the following thermochemical data, what is the change in enthalpy for the following reaction?

Answer 1

Answer:

D. -120.9 kJ

Explanation:

According to Hess's law ,the total enthalpy change for a reaction is the sum of all changes regardless of the stages or the steps of the reaction.

$CaO + 2HCl \rightarrow CaCl_{2} + H_{2}O\ (\Delta H = -186\ kJ)$....(1)

$CaO + H_{2}O\rightarrow Ca(OH)_{2}\ (\Delta H = - 65 \ kJ)$

(this reaction should be reversed in order to reach the required reaction )

On reversing the reaction the sign of $\Delta H$ get reversed.

(In this case change sign from '-' to'+'. Hence  $\Delta H$ = + 65 kJ)

$CaO + 2HCl \rightarrow CaCl_{2} + H_{2}O\ (\Delta H = - 186\ kJ)$....(1)

$Ca(OH)_{2} \rightarrow CaO + H_{2}O\ (\Delta H = + 65 kJ )$......(2)

Adding equation (1) and (2)

$Ca(OH)_{2} + 2HCl \rightarrow CaCl_{2} + 2H_{2}O[tex][tex]Delta H = - 186 + 65 = - 121\kJ$

$Delta H = - 121\kJ$ (It is nearly equal to -120.9 kJ)