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For a chemical reaction, 4A+3B→6C+9D4A + 3B \rightarrow 6C + 9D rate of formation of C is 6×10–2 mol L–1 s–16 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and rate of disappearance of A is 4×10–2 mol L–1 s–14 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}. The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be:

A

1×10–2 mol L–1 s–11 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 30×10–2 mol L–130 \times 10^{–2} \text{ mol L}^{–1}

B

10×10–2 mol L–1 s–110 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 10×10–2 mol L–110 \times 10^{–2} \text{ mol L}^{–1}

C

1×10–2 mol L–1 s–11 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 10×10–2 mol L–110 \times 10^{–2} \text{ mol L}^{–1}

D

10×10–2 mol L–1 s–110 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 30×10–2 mol L–130 \times 10^{–2} \text{ mol L}^{–1}

Step-by-Step Solution

For the given reaction: 4A+3B→6C+9D4A + 3B \rightarrow 6C + 9D The rate of reaction is related to the rates of individual components as follows: Rate=−14d[A]dt=−13d[B]dt=16d[C]dt=19d[D]dt\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, d[C]dt=6×10−2 mol L−1 s−1\frac{d[C]}{dt} = 6 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1} Rate of disappearance of A, −d[A]dt=4×10−2 mol L−1 s−1-\frac{d[A]}{dt} = 4 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}

Rate of reaction =16×d[C]dt=16×(6×10−2)=1×10−2 mol L−1 s−1= \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 =−d[B]dt=3×Rate of reaction=3×(1×10−2)=3×10−2 mol L−1 s−1= -\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 Δt=10 s\Delta t = 10 \text{ s} is: Δ[B]=(−d[B]dt)×Δt=(3×10−2 mol L−1 s−1)×10 s=30×10−2 mol L−1\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}

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