For the given reaction: $4A + 3B \rightarrow 6C + 9D$ The rate of reaction is related to the rates of individual components as follows: $\text{Rate} = -\frac{1}{4}\frac{d[A]}{dt} = -\frac{1}{3}\frac{d[B]}{dt} = \frac{1}{6}\frac{d[C]}{dt} = \frac{1}{9}\frac{d[D]}{dt}$ Given: Rate of formation of C, $\frac{d[C]}{dt} = 6 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$ Rate of disappearance of A, $-\frac{d[A]}{dt} = 4 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$

Rate of reaction $= \frac{1}{6} \times \frac{d[C]}{dt} = \frac{1}{6} \times (6 \times 10^{-2}) = 1 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$

Rate of disappearance of B $= -\frac{d[B]}{dt} = 3 \times \text{Rate of reaction} = 3 \times (1 \times 10^{-2}) = 3 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}$

The amount of B consumed in an interval of $\Delta t = 10 \text{ s}$ is: $\Delta [B] = \left(-\frac{d[B]}{dt}\right) \times \Delta t = (3 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}) \times 10 \text{ s} = 30 \times 10^{-2} \text{ mol L}^{-1}$