N_2 + 3H_2 to 2NH_3. If the concentration of | vedclass

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(B) The rate of formation of $NH_3$ is given by: $\frac{d[NH_3]}{dt} = \frac{\Delta [NH_3]}{\Delta t} = \frac{0.04 \ M - 0.01 \ M}{20 \ s} = \frac{0.0

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$0.75 	imes 10^{-3} \ M \ sec^{-1}$

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(B) The rate of formation of $NH_3$ is given by: $\frac{d[NH_3]}{dt} = \frac{\Delta [NH_3]}{\Delta t} = \frac{0.04 \ M - 0.01 \ M}{20 \ s} = \frac{0.03 \ M}{20 \ s} = 1.5 	imes 10^{-3} \ M \ s^{-1}$.For the reaction $N_2 + 3H_2 	o 2NH_3$,the rate of reaction $(r)$ is defined as: $r = \frac{1}{2} \frac{d[NH_3]}{dt}$.Substituting the value: $r = \frac{1}{2} 	imes 1.5 	imes 10^{-3} \ M \ s^{-1} = 0.75 	imes 10^{-3} \ M \ s^{-1}$.

$N_2 + 3H_2 \to 2NH_3$. If the concentration of $NH_3$ changes from $0.01 \ M$ to $0.04 \ M$ in $20 \ seconds$,then the rate of reaction will be:

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