Question

Calculate the standard enthalpy of formation of liquid benzene given enthalpies carbon hydrogen benzene -393.5kj , - 285.83kj, and 3267.0kj

Answer 1

Answer:

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Answer 2

GIVEN :

$C + O_2 \rightarrow CO_2$       $\Delta_C H = - 393.5kJ$

$H_2 +\frac{1}{2} O_2 \rightarrow H_2O$   $\Delta_C H = - 285.83 kJ$

$C_6H_6 +\frac{15}{2} O_2 \rightarrow 3H_2O + 6CO_2$   $\Delta_C H = 3267 kJ$

To find:

Standard enthalpy of formation of benzene

Explanation:

Reversing the third equation given :

$3H_2O + 6CO_2 \rightarrow C_6H_6 +\frac{15}{2} O_2$     $\Delta_C H = 3267 kJ$       $\rightarrow 1$

Multiplying the first equation with 6:

$6C + 6O_2 \rightarrow 6CO_2$                          $\Delta_C H = -2361 kJ$      $\rightarrow 2$

Multiplying second equation with 3:

$3H_2 +\frac{3}{2} O_2 \rightarrow 3H_2O$                       $\Delta_C H = -857.49$$kJ$    $\rightarrow 3$

Rearranging the equations to obtain the equation for benzene formation:

$1 + 2 + 3$ $\rightarrow$   $6C + 3H_2 \rightarrow C_6H_6$  

$\Delta_fH = 3267 kJ - 2361 kJ - 857.49kJ$ $= 48.51 kJ$

Therefore enthalpy of formation of liquid benzene $= 48.51kJ$