Question

For the following equilibrium: A +2B⟺C+3D If the change in concentration for B (ΔB) is 0.44 M, what will be the change in concentration for C (ΔC)? (Round your answer to two decimal places)

Answer 1

The change in concentration for C (ΔC) is half of the change in concentration for B (ΔB) due to the stoichiometry of the reaction, which gives a value of 0.22 M when rounded to two decimal places.

Step-by-step explanation:

For the equilibrium A + 2B ⇄ C + 3D, if the change in concentration for B (ΔB) is 0.44 M, the change in concentration for C (ΔC) can be determined by using the stoichiometry of the reaction. Since A reacts with 2B to produce C, the ratio of the change in B to the change in C is 2:1. This means that for every 2 moles of B reacting, 1 mole of C is produced.

Therefore, if ΔB is 0.44 M, ΔC would be half of that, because of the 1:2 stoichiometric ratio between B and C in the reaction. The change in concentration for C (ΔC) would then be:

ΔC = ΔB / 2 = 0.44 M / 2

ΔC = 0.22 M

Thus, the change in concentration for C (ΔC) after the equilibrium shift will be 0.22 M, rounded to two decimal places.