Question

Consider the following chemical equations to answer the question that follows. CaCO3(s)Ca(OH)2(s)Ca(OH)2(s)+2HCl(g)→CaO(s)+CO2(g)→H2O(l)+CaO(s)→CaCl2(s)+2H2O(l)ΔHΔHΔH=175kJ=67kJ=−198kJUsing the information above, determine the change in enthalpy for the following chemical reaction.CaCO3(s)+2HCl(g)⟶CaCl2(s)+H2O(l)+CO2(g)

Answer 1

Answer: -90 kJ

Step-by-step explanation:

Hess's Law states that if a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. If we reverse the second reaction (giving us ΔH=−67 kJ), we can then write the sum of three equations above as follows.

CaCO3(s) --> CaO(s)+CO2(g)

H2O(l)+CaO(s) --> Ca(OH)2(s)

Ca(OH)2(s)+2HCl(g) --> CaCl2(s)+2H2O(l)

------------------------------------------------------------

CaCO3(s)+2HCl(g) ---> CaCl2(s)+H2O(l)+CO2(g)

Therefore, the change in enthalpy is calculated as follows.

ΔH = (175kJ)+(−67kJ)+(−198kJ)

= −90.kJ

Answer 2

Answer : The change in enthalpy for the following chemical reaction is, -90 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

`CaCO_3(s)+2HCl(g)\rightarrow CaCl_2(s)+H_2O(l)+CO_2(g)`
`\Delta H=?`

The intermediate balanced chemical reaction will be,

(1)
`CaCO_3(s)\rightarrow CaO(s)+CO_2(g)`
`\Delta H_1=175kJ`

(2)
`Ca(OH)_2(s)\rightarrow H_2O(l)+CaO(s)`
`\Delta H_2=67kJ`

(3)
`Ca(OH)_2(s)+2HCl(g)\rightarrow CaCl_2(s)+2H_2O(l)`
`\Delta H_3=-198kJ`

Now we will reverse the reaction 2 and then adding all the equations, we get :

(1)
`CaCO_3(s)\rightarrow CaO(s)+CO_2(g)`
`\Delta H_1=175kJ`

(2)
`H_2O(l)+CaO(s)\rightarrow Ca(OH)_2(s)`
`\Delta H_2=-67kJ`

(3)
`Ca(OH)_2(s)+2HCl(g)\rightarrow CaCl_2(s)+2H_2O(l)`
`\Delta H_3=-198kJ`

The expression for enthalpy of change will be,

`\Delta H=\Delta H_1+\Delta H_2+\Delta H_3`

`\Delta H=(175kJ)+(-67kJ)+(-198kJ)`

`\Delta H=-90kJ`

Therefore, the change in enthalpy for the following chemical reaction is, -90 kJ