Question

1. Consider the reaction: J2(g) + R2(g) ⇄ 2JR(g). Let a square represent an atom of J and a circle represent an atom of R. The value for the equilibrium constant for this reaction at 25°C is greater than 1. The drawing below labeled Before shows a small representative volume of a container held at 25°C immediately after J2 and R2 have been added but before reaction between J2 and R2 to give JR has occurred. Circle the letter of each choice below that would be a possible drawing of the same small representative volume after equilibrium has been attained in the system described by the chemical equation above. For each incorrect choice, state why it is incorrect; be as specific as possible.

Before COo 0 O Co None of these Oo o

Answer 1

It is observed that one molecule of J2 reacts with one molecule of R2 to give two molecules of JR. The equilibrium constant is greater than 1, that means, at equilibrium, there will be more product molecules than reactants. On basis of this facts, the answer of the question is E. Detailed explanations are as follows.

It is observed that one molecule of J2 reacts with one molecule of R2 to give two molecules of JR. The equilibrium constant is greater than 1, that means, at equilibrium, there will be more product molecules than reactants. On basis of this facts, the answer of the question is E. Detailed explanations are as follows. J2 (g) + R_2 (g) rightharpoonoverleftharpoon 2 JR (g) It can be concluded that 1 molecule of J_2 reacts with 1 molecule of R_2 to give 2 molecules of JR. Now, initially in the box (small position of container) there was 3 molecules of J_2 and 5 molecules of R_2 (infinity) present . From the above equation, we can say that 3 molecules of J_2 will react with 3 R_2 molecules to give (3 times 2) = 6 molecules of JR But since the system is in equilibrium some JR will decompose to give back J_2 and R_2 i.e J_2 or R_2 will not be finished at equilibrium.