In the reaction 2N_2O_5(g) rightarrow 4 NO_2( | vedclass

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(B) The rate of reaction is given by the expression:Rate $= -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$Giv

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$0.03 	ext{ mol dm}^{-3} 	ext{s}^{-1}$

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(B) The rate of reaction is given by the expression:Rate $= -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$Given that the rate of disappearance of $N_2O_5$ is $-\frac{d[N_2O_5]}{dt} = 0.06 	ext{ mol dm}^{-3} 	ext{s}^{-1}$.Substituting this into the rate expression:Rate $= \frac{1}{2} 	imes (0.06 	ext{ mol dm}^{-3} 	ext{s}^{-1}) = 0.03 	ext{ mol dm}^{-3} 	ext{s}^{-1}$.Since the rate of formation of $O_2$ is equal to the rate of reaction:Rate of formation of $O_2 = 0.03 	ext{ mol dm}^{-3} 	ext{s}^{-1}$.

In the reaction $2N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)$, $N_2O_5$ disappears at a rate of $0.06 \text{ mol dm}^{-3} \text{s}^{-1}$. Calculate the rate of formation of $O_2(g)$.

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