Consider the reaction CO2(g)+ H2(g)- CO(g) H20(g) at 25°C (298.15 K). This water

Question

Consider the reaction CO2(g)+ H2(g)- CO(g) H20(g) at 25°C (298.15 K). This water-gas-shift reaction is commonly encountered in the chemical industry, though it takes place only at temperatures well above 25°C (298.15K) when water is a vapor. However, the data used are for 25°C (298.15K) when water is a liquid. Calculate the enthalpy change for the physical change that transforms water from its standard state as a liquid into its standard state as an ideal gas. Use the answer from part a) to calculate the standard heat of reaction at 25°C (298.15K) a. b.

Explanation / Answer

Part a

Formation of CO2 from its elements

C(s) + O2(g) = CO2(g)

Hf = - 393.5 kJ/mol

Reverse the above reaction

CO2(g) = C(s) + O2(g)...........1

H1 = 393.5 kJ/mol

Formation of CO from its elements

C(s) + 0.5O2(g) = CO(g) .........2

H2 = - 110.5 kJ/mol

Formation of H2O from its elements

H2(g) + 0.5O2(g) = H2O(g)....... 3

H3 = - 241.8 kJ/mol

Add 1, 2 and 3

CO2(g) + H2(g) = CO(g) + H2O (g)

Heat of reaction

H = H1 + H2 + H3

= 393.5 - 110.5 - 241.8

= 41.2 kJ/mol

= 41200 J/mol

Heat of formation of H2O (l) = 285.8 kJ/mol

Enthalpy change for

H2O(l) = H2O(g)

H' = Hf(H2O g) - Hf(H2O l)

= - 241.8 - (-285.8)

= 44 kJ/mol

= 44000 J/mol

Part b

CO2(g) = C(s) + O2(g)...........1

H1 = 393.5 kJ/mol

Formation of CO from its elements

C(s) + 0.5O2(g) = CO(g) .........2

H2 = - 110.5 kJ/mol

Formation of H2O(l) from its elements

H2(g) + 0.5O2(g) = H2O(l)....... 3

H3 = - 285.8 kJ/mol

Formation of H2O(g)

H2O(l) = H2O(g)............ 4

H' = 44 kJ/mol

Now add 1, 2, 3 and 4

CO2(g) + H2(g) = CO(g) + H2O (g)

Heat of reaction

H = H1 + H2 + H3 + H'

= 393.5 - 110.5 - 285.8 + 44

= 41.2 kJ/mol

= 41200 J/mol

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