Question

Consider the following reaction occurring in a closed chemical system. Assume that this reaction is at equilibrium.2C4H10 + 13O2 <-> 8CO2 + 10H2O∆H = –2,877.5 kJ/mol(a)Is this an endothermic or exothermic reaction? Explain your reasoning.(b)If C4H10 is removed from the system, how will the reactions shift to reach equilibrium again? Explain.(c)If water is added to the system, how will the reactions shift to reach equilibrium again? Explain.(d)If heat is added to the system, how will the reactions shift to reach equilibrium again? Explain.(e)What is the name of the principle that helps you predict each of these shifts in equilibrium?

Answer 1

2C4H10 + 13O2 <-> 8CO2 + 10H2O∆H = –2,877.5 kJ/mol

a) This is an exothermic reaction, you can realize because of the ∆H. The sign is negative so it releases heat. It is the enthalpy of the reaction.

b) If you remove C4H10 then the equilibrium will shift to the left. The amount of reactant that was removed must be restored to get again the equilibrium.

c) Remember the reaction is exothermic so it releases heat.

You can see this as we did in b) ok?

Think this: if you remove reactants it goes to the left, if you remove products it goes to the right.

Now think if we add reactants, it will shift to the right. If we add products will shift to the left.

Now If you have an Exo Reaction, the heat release you can see it as a product of the reaction. So, if you add heat, then it will shift to the left!

Remember we have an Exothermic reaction.

d) This is Le Chatelier's principle.