Question

Calculate delta h for the reaction c2h4 g h2 g c2h6 g

Answer 1

calculate enthalpy change foe the reaction -:C2H4(g) +H2(g) ->C2H6(g) using the following combustion data-: C2H4(g) +3O2(g) ->2CO2(g) +2H2O(l) ;ΔH0 =-140KJ/MOL. C2H6(g) +​7/2O2(g)->2CO2(g) + 3H2O(l) ;ΔH0=-1550KJ/MOL.

Answer 2

Answer: The enthalpy of the reaction is coming out to be -136.28 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(C_2H_6)})]-[(1\times \Delta H_f_{(H_2)})+(1\times \Delta H_f_{(C_2H_4)})]$

We are given:

$\Delta H_f_{(C_2H_4)}=52.53kJ/mol\\\Delta H_f_{(C_2H_6)}=-83.75kJ/mol\\\Delta H_f_{(H_2)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-83.75))]-[(1\times (52.53))+(1\times (0))]\\\\\Delta H_{rxn}=-136.28kJ$

Hence, the enthalpy of the reaction is coming out to be -136.28 kJ.