Question

Problem Solving 11) Determine the reaction order and rate constant in each run then calculate the average rate constant, from the following data. (10 points) A + B C + D Sample [A] 1.606x10" 1.606x10" 1.071x100 [B] Rate 1.006x10" 8.70x10 6.710x1026.40x10 1.006x10° 4.30x10" -d[A]/dt = k[A]"[B]"

Answer 1

Please provide the entire details for question 11.

For question 13,the answer is as follows

Given are the following information in the question

C₂H₆ + 7/2 O₂ $\rightarrow$ 2CO₂ + 3H₂O  $\Delta$H = -78kJ..................(1)

C₂H₄+ 5/2O₂ $\rightarrow$ 2CO₂ + 2H₂O $\Delta$ H = -285.83kJ...................(2)

We need $\Delta$ H of the equation C₂H₄ + H₂$\rightarrow$ C₂H₆

Reversing equation (1) we get

2CO₂ + 3H₂O $\rightarrow$ C₂H₆ + 7/2 O₂ $\Delta$ H = 78kJ..................(3)

Adding equation (3) and equation (2) , we get

C₂H₄+ 5/2O₂ + 2CO₂ + 3H₂O $\rightarrow$ 2CO₂ + 2H₂O + C₂H₆ + 7/2 O₂$\Delta$H = ( -285.83 + 78) kJ

Simplifying the above equation,we get

C₂H₄+ 5/2O₂ + 3H₂O $\rightarrow$ 2H₂O + C₂H₆ + 7/2 O₂ $\Delta$ H = ( -207.83) kJ

_{$\Rightarrow$}C₂H₄+ 5/2O₂ + H₂O $\rightarrow$   C₂H₆ + 7/2 O₂

_{$\Rightarrow$}C₂H₄+ H₂ + O $\rightarrow$   C₂H₆ + O

Finally, we get the equation for formation of ethane from ethene , C₂H₄+ H₂ $\rightarrow$   C₂H₆ ....(4)   whose $\Delta$ H = -207.83kJ

Thus, the enthalpy of ethane obtained from the reaction of ethene and hydrogen is -207.83kJ and the three equations at equilibrium are (1) ,(2) and (4).