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

Answer:

The enthalpy for given reaction is 232 kilo Joules.

Explanation:

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

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

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

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

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

$\Delta H^o_{4}=2\times \Delta H^o_{2}+\Delta H^o_{3} - \Delta H^o_{1}$

$\Delta H^o_{4}=2\times (-340 kJ) + (-211 kJ) - (-1,123 kJ)$

$\Delta H^o_{4}=232 kJ$

The enthalpy for given reaction is 232 kilo Joules.