Question

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

A.

120.9 kJ

B.

-251.1 kJ

C.

251.1 kJ

D.

-120.9 kJ

Answer 1

D. -120.9 kJ

Step-by-step 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.

`<strong>CaO + 2HCl \rightarrow CaCl_(2) + H_(2)O\ (</strong>\<strong>Delta H = -186\</strong> kJ)`....(1)

`<strong>CaO + H_(2)O\rightarrow Ca(OH)_(2)\ </strong>(\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)

`<strong>CaO + 2HCl \rightarrow  CaCl_(2) + H_(2)O\ (</strong>\<strong>Delta H = - 186\</strong> kJ)`....(1)

`<strong>Ca(OH)_(2)   \rightarrow  CaO + H_(2)O\</strong> (\<strong>Delta H</strong> = <strong>+ 65 kJ</strong> )`......(2)

Adding equation (1) and (2)

`<strong>Ca(OH)_(2) + 2HCl \rightarrow CaCl_(2) + 2H_(2)O</strong>[tex]</p><p>[tex]<strong>Delta H = - 186 + 65 = - 121\kJ</strong>`

`<strong>Delta H = - 121\kJ</strong>` (It is nearly equal to -120.9 kJ)