These are four questions and four answers.
Answers:
1) activation energy of the reverse reaction
b. Decreased
2) Rate of the forward reaction
a. Increased
3) Rate of the reverse reaction
a. Increased
4) Activation energy of the forward reaction
b. decreased
Explanation:
<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.
<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.
The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate. The equlibrium reactions may be represented by:
Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).
For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.
Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:
1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)
2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)
3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).
4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).
In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.
(where Q is heat, m is mass, c is specific heat and 
is change in temperature. So according this formula, increasing mass will increase the substance's heat, but won't effect it's temperature since they are not related. Unless, if you want to keep the substance's heat constant, in that case when you increase it's mass you will have to decrease the temperature