Question

Question 12 (3 points) Find AH for C4H10 (g) + 8 H20 (g) 4CO2 (g) + 13 H2 given C4H10 (g) + 4H20 (g) 4 CO (g) + 9 H2(g) AH=651 kJ CO2 (g) + H2(g) →CO (g) + H20 (g) AH = 41 kJ (SHOW CALCULATIONS) DSC . Format in E h Question 13 (4 points) How many moles of Cl2 are in a 200.0 mL container at 610 torr and 135°C? (SHOW CALCULATIONS) DC - Format B I U -

Answer 1

The given reaction is

   C4H10 (g) + 8 H2O (g) $\rightarrow$ 4 CO2 (g) + 13 H2    $\Delta$ H = ?   --------------------- (1)

According to Hess law the the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states. That is if a chemical change takes place by several different routes, the overall enthalpy change is the same, regardless of the route by which the chemical change occurs.

otherwise in simple manner for applications we can say like this,

regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

we are going to apply Hess Law in this problem.

Consider the other two given reactions

C4H10 (g) + 4 H2O (g) $\rightarrow$ 4 CO (g) + 9 H2 (g)   $\Delta$ H = 651 KJ ----------------- (2)

    CO2 (g) + H2(g)   $\rightarrow$ CO (g) + H2O (g)    $\Delta$ H = 41 KJ -------------------(3)

rewrite equation (3) in reverse [ note that the $\Delta$ H will change the sign while reversing the reaction]

CO (g) + H2O (g) $\rightarrow$ CO2 (g) + H2(g)     $\Delta$ H = - 41 KJ -------------------(4)

Now multiply the reaction (4) by 4 and add to reaction (2) to get the reaction (1)

4CO (g) + 4 H2O (g) $\rightarrow$ 4 CO2 (g) + 4 H2O (g)   $\Delta$ H = -164 KJ

C4H10 (g) + 4 H2O (g) $\rightarrow$ 4 CO (g) + 9 H2 (g)   $\Delta$ H = 651 KJ

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Adding

   C4H10 (g) + 8 H2O (g) $\rightarrow$ 4 CO2 (g) + 13 H2    $\Delta$ H = -164 + 651 = 487 KJ

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Hence the $\Delta$ H for the required reaction is 487 KJ