Question

3. Adjust the temperature for an endothermic reaction. Then adjust the temperature for an exothermic reaction. What happens? K... When I decreases creases Sign of AH When Tincreases Positive (endothermic) Negative (exothermic) a. Write the chemical equation for the endothermic reaction when An=0. Is heat a reactant or a product? b. Write the chemical equation for the exothermic reaction when An=0. Is heat a reactant or a product? C. How does an endothermic reaction shift to regain equilibrium after an increase in temperature? Explain this in terms of Q versus K and also in terms of "reducing stress" after a change in a reactant or product.

can someone help me explain how to answer these question using this simulation ?

Answer 1

Answer 3. Adjust the temperature for an endothermic reaction.

The effect of temperature on endothermic reaction is (by applying Le Chatelier's principle) has to do with the heat of reaction. So if we raise the temperature on an endothermic reaction, it is like we are adding more reactant and therefore by applying Le Chatelier's principle the equillibrium will shift towards the right (i.e. towards product side) and the reaction will be feasible and spontaneous. Similarly, in converse, if we decrease the temperature on an endothermic reaction, it is like we are removing reactant and therefore by applying Le Chatelier's principle the equillibrium will shift towards the left (i.e. towards reactant side) and the reaction will not be feasible and spontaneous.

Then adjust the temperature for an exothermic reaction. The effect of temperature on exothermic reaction is (by applying Le Chatelier's principle) has to do with the heat of reaction. So if we raise the temperature on an exothermic reaction, it is like we are adding more product and therefore by applying Le Chatelier's principle the equillibrium will shift towards the left (i.e. towards reactant side) and the reaction will not be feasible and spontaneous. Similarly, in converse, if we decrease the temperature on an exothermic reaction, it is like we are removing product and therefore by applying Le Chatelier's principle the equillibrium will shift towards the right (i.e. towards product side) and the reaction will be feasible and spontaneous.

The effect of increasing/decreasing temperature on the value of equillibrium contant (K) and sign of $\Delta$ H is as follows:

sign of $\Delta$ H when T increases K---when T decreases K---- when T increases

Positive (endothermic) Decreases (endothermic) Increases (endothermic)

Negative (exothermic) Increases (exothermic) Decreases (exothermic)

Answer 3a. The chemical equation for the endothermic reaction when $\Delta$ n = 0 is as follows:

Heat + A $\rightleftharpoons$ B $\Delta$ H = +

As we see from the above equation heat is reactant in endothermic reaction.

Answer 3b.

The chemical equation for the exothermic reaction when $\Delta$ n = 0 is as follows:

A $\rightleftharpoons$ B + Heat   $\Delta$H = -

As we see from the above equation heat is product in exothermic reaction.

Answer 3c. The endothermic reaction shift as follows to regain equillibrium after an increase in temperature. This we will explain in terms of Q versus K and also in terms of 'reducing stress' after a change in reactant or product.

Now let us first define Q i.e. reaction quotient that expresses the relative ratio of product to reactant at a given instant and K is equillibrium constant. Now from Q versus K we can compare and then there is three situation

(1) If K Q then the reaction will proceed forward converting reactant into preoduct.

(2) if K Q then the reaction will proceed in the reverse direction i.e. converting produycts into reactant.

(3) If K = Q then the system is at equillibrium.

Now considering the above fact in endothermic reaction on increasing the temperature K Q and so the reaction will proceed forward converting reactant into product and after certain level of increase in temperature K = Q i.e. K will become equal to Q and now reaction will be in equillibrium state.

If there is further increase in temperature then K Q and the reaction will proceed in the reverse direction i.e. converting products into reactant and equillibrium will be again changed.

For endothermic reaction heat relieves the stress and reduces it. So on increasing the temperature (i.e. heat i.e. reactant in endothermic reaction) stress will be reduced and the reaction will proceed spontaneously in right direction i.e. converting reactant into product. That will occur upto the stage at which reaction reaches equillibrium state.

Answer 3d. The exothermic reaction shift as follows to regain equillibrium after an increase in temperature. This we will explain in terms of Q versus K and also in terms of 'reducing stress' after a change in reactant or product.

Now let us first define Q i.e. reaction quotient that expresses the relative ratio of product to reactant at a given instant and K is equillibrium constant. Now from Q versus K we can compare and then there is three situation

(1) If K Q then the reaction will proceed forward converting reactant into preoduct.

(2) if K Q then the reaction will proceed in the reverse direction i.e. converting produycts into reactant.

(3) If K = Q then the system is at equillibrium.

Now considering the above fact in exothermic reaction on increasing the temperature K Q and so the reaction will proceed in the reverse direction  converting product into reactant and so in exothermic reaction on increasing the temperature the equillibrium position will be towards left i.e. reactant side. Now if we decrease the temperature then K Q and then the reaction will proceed forward converting reactant into product and a stage is reached when K= Q i.e. reaction reaches equillibrium state.

For exothermic reaction heat add the stress to the reaction system. So on increasing the temperature (i.e. heat i.e. product in exothermic reaction) stress will be increased and the reaction will proceed spontaneously in left direction i.e. converting product into reactant. So we have to remove the stress i.e. heat by decreasing the temperature to let the reaction proceed in the right direction i.e. converting reactant into product upto the reaction reaches equillibrium.