For the reaction 2A ⇌ B + C, if the initial concentration of A is 0.5 M and at equilibrium, the concentration of B is 0.2 M, what is the equilibrium concentration of A?
Practice Questions
1 question
Q1
For the reaction 2A ⇌ B + C, if the initial concentration of A is 0.5 M and at equilibrium, the concentration of B is 0.2 M, what is the equilibrium concentration of A?
0.1 M
0.2 M
0.3 M
0.4 M
At equilibrium, [A] = 0.5 - 0.2 = 0.3 M.
Questions & Step-by-step Solutions
1 item
Q
Q: For the reaction 2A ⇌ B + C, if the initial concentration of A is 0.5 M and at equilibrium, the concentration of B is 0.2 M, what is the equilibrium concentration of A?
Solution: At equilibrium, [A] = 0.5 - 0.2 = 0.3 M.
Steps: 5
Step 1: Write down the initial concentration of A, which is 0.5 M.
Step 2: Identify the change in concentration of A when the reaction reaches equilibrium. Since 2 moles of A produce 1 mole of B, for every 1 mole of B produced, 2 moles of A are consumed.
Step 3: At equilibrium, the concentration of B is given as 0.2 M. This means that 0.2 M of B was produced.
Step 4: Calculate how much A was consumed to produce 0.2 M of B. Since 2 moles of A are needed to produce 1 mole of B, we need 2 * 0.2 M = 0.4 M of A.
Step 5: Subtract the amount of A consumed from the initial concentration of A. So, [A] at equilibrium = 0.5 M (initial) - 0.4 M (consumed) = 0.1 M.
