Question

C(graphite) + o2(g) → co2(g)δh o rxn = −393.5 kj/mol h2(g) + 1 2 o2(g) → h2o(l)δh o rxn = −285.8 kj/mol 2c2h6(g) + 7o2(g) → 4co2(g) + 6h2o(l)δh o rxn = −3119.6 kj/mol calculate the enthalpy change for the reaction below: 2 c(graphite) + 3h2(g) → c2h6(g)

Answer 1

Answer : The enthalpy change for the reaction, $\Delta H_{rxn}=-85KJ$

Solution : Given,

(1) $C(graphite)+O_2(g)\rightarrow CO_2(g)$         $\Delta H_1=-393.5KJ/mole$

(2) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$      $\Delta H_2=-285.8KJ/mole$

(3) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$     $\Delta H_3=-3119.6KJ/mole$

$2C(graphite)+3H_2(g)\rightarrow C_2H_6(g)$       $\Delta H_{rxn}=?$

Now we have to calculate the enthalpy change for this reaction by doing:

$2\times eq.(1)+\frac{1}{2}(eq.3)\text{[reversing equation 3 and dividing it by 2]}+3(eq.2)$

$\Delta H_{rxn}=2\times \Delta H_1+3\times \Delta H_2+(-\frac{\Delta H_3}{2})$

$\Delta H_{rxn}=2\times (-393.5)+3\times (-285.8)+(-\frac{-3119.6}{2})$

$\Delta H_{rxn}=-85KJ$

Therefore, the enthalpy change for the reaction, $\Delta H_{rxn}=-85KJ$

Answer 2

The enthalpy for the change in reaction will be -85 kJ.

In the equation,

Decomposition of graphite = -393.5 kJ/mol

enthalpy for formation of $\rm H_2O$ = -285.8 kJ/mol

decomposition of $\rm C_2H_6$ = -3119.6 kJ/mol

The enthalpy for reaction of graphite with hydrogen will be:

=  2 $\times$ Decomposition of graphite + 3 $\times$ enthalpy for formation of $\rm H_2O$ - $\rm \frac{decomposition\;of\;C_2H_6 }{2}$

= 2 $\times$ -395.5 + 3 $\times$ -285.8 - $\rm \frac{-3119.6}{2}$ kJ

= -85 kJ

The enthalpy for the change in reaction will be -85 kJ.

For more information about reaction enthalpy, refer the link:

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