Question

Calculate the enthalpy for this reaction: 2C(s) + H2(g) ---> C2H2(g) ΔH° = ??? kJ Given the following thermochemical equations: C2H2(g) + (5/2)O2(g) ---> 2CO2(g) + H2O(ℓ) ΔH° = -1,123 kJ C(s) + O2(g) ---> CO2(g) ΔH° = -340 kJ H2(g) + (1/2)O2(g) ---> H2O(ℓ) ΔH° = -211 kJ

Answer 1

The enthalpy for given reaction is 232 kilo Joules.

Step-by-step explanation:

`C_2H_2(g) + (5)/(2)O_2(g)\rightarrow 2CO_2(g) + H_2O(l), \Delta H^o_(1) = -1,123 kJ`...[1]

`C(s) + O_2(g)\rightarrow CO2(g), \Delta H^o_(2) = -340 kJ`..[2]

`H_2(g) + (1)/(2)O_2(g)\rightarrow H_2O(l) ,\Delta H^o_(3) = -211 kJ`..[3]

`2C(s) + H_2(g)\rightarrow C_2H_2(g),\Delta H^o_(4) =?`..[4]

2 × [2] + [3] - [1] ( Using Hess's law)

`\Delta H^o_(4)=2* \Delta H^o_(2)+\Delta H^o_(3) - \Delta H^o_(1)`

`\Delta H^o_(4)=2* (-340 kJ) + (-211 kJ) - (-1,123 kJ)`

`\Delta H^o_(4)=232 kJ`

The enthalpy for given reaction is 232 kilo Joules.