Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration of A was measured to be -0.0800 M/s. What is the final concentration of C at the end of this same interval if its concentration was initially 3.000 M? 3.960 M

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B > C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

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