Answer:
rA = 0.60 M/s
rC = 0.90 M/s
Explanation:
Let's consider the following reaction:
2 A+B ⇒ 3 C
The rate of each substance can be calculated like the change in its concentration divided by the change in time. Given the rate must always be positive, we add a minus sign before the reactants change in concentration.
[tex]rA=-\frac{\Delta[A] }{\Delta t}[/tex]
[tex]rB=-\frac{\Delta[B] }{\Delta t}[/tex]
[tex]rC=\frac{\Delta[C] }{\Delta t}[/tex]
The rate of the reaction is equal to the rate of each substance divided by its stoichiometric coefficient.
[tex]r= \frac{rA}{2} =\frac{rB}{1} =\frac{rC}{3}[/tex]
The rate of disappearance of B is 0.30 M/s.
The rate of disappearance of A is:
[tex]\frac{rA}{2} =\frac{rB}{1}\\rA = 2 \times rB = 2 \times 0.30 M/s = 0.60 M/s[/tex]
The rate of appearance of C is:
[tex]\frac{rB}{1} =\frac{rC}{3}\\rC = 3 \times rB = 3 \times 0.30 M/s = 0.90 M/s[/tex]
