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Chemistry Question on Chemical Kinetics

Decomposition of H2O2H_2O_2 follows a first order reaction. In fifty minutes the concentration of H2O2H_2O_2 decreases from 0.50.5 to 0.125 M0.125 \,M in one such decomposition. When the concentration of H2O2H_2O_2 reaches 0.05 M0.05 \,M, the rate of formation of O2O_2 will be :

A

6.93×10−2 mol min−16.93 \times 10^{-2} \,mol \, min^{-1}

B

6.93×10−4 mol min−16.93 \times 10^{-4} \,mol \, min^{-1}

C

2.66 L min−12.66 \,L \, min^{-1} at STPSTP

D

1.34×10−2mol min−11.34 \times 10^{-2} mol \, min^{-1}

Answer

6.93×10−4 mol min−16.93 \times 10^{-4} \,mol \, min^{-1}

Explanation

Solution

t3/4=2×t1/2=50min⁡t_{3/4} = 2 \times t_{1/2} = 50 \min

i.e. t1/2=25min⁡t_{1/2} = 25 \min
k=0.693t1/2=0.69325min⁡−1k = \frac{0.693}{t_{1/2}} = \frac{0.693}{25} \min^{-1}

Rate of H2O2H_2O_2 decomposition =k[H2O2]= k[H_{2}O_{2}]
=0.69325×0.05=−d[H2O2]dt= \frac{0.693}{25} \times0.05 = - \frac{d\left[H_{2}O_{2}\right]}{dt}
H2O2⟶H2O+12O2H _{2} O _{2} \longrightarrow H _{2} O +\frac{1}{2} O _{2}
−d[H2O2]dt=2d[O2]dt- \frac{d\left[H_{2}O_{2}\right]}{dt} = 2 \frac{d\left[O_{2}\right]}{dt}
⇒d[O2]dt=6.93×10−4molmin⁡−1\Rightarrow \frac{d\left[O_{2}\right]}{dt} = 6.93 \times10^{-4} \text{mol} \min^{-1}