For the reaction N_2 + 3H_2 to 2NH_3,if frac{ | vedclass

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(B) The rate of reaction for $N_2 + 3H_2 	o 2NH_3$ is given by: $-\frac{\Delta [N_2]}{\Delta t} = -\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{

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$3 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$

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(B) The rate of reaction for $N_2 + 3H_2 	o 2NH_3$ is given by: $-\frac{\Delta [N_2]}{\Delta t} = -\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [NH_3]}{\Delta t}$ From the relation $-\frac{1}{3} \frac{\Delta [H_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [NH_3]}{\Delta t}$,we get: $-\frac{\Delta [H_2]}{\Delta t} = \frac{3}{2} 	imes \frac{\Delta [NH_3]}{\Delta t}$ Substituting the given value $\frac{\Delta [NH_3]}{\Delta t} = 2 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$: $-\frac{\Delta [H_2]}{\Delta t} = \frac{3}{2} 	imes 2 	imes 10^{-4} \ mol \ L^{-1} s^{-1} = 3 	imes 10^{-4} \ mol \ L^{-1} s^{-1}$

For the reaction $N_2 + 3H_2 \to 2NH_3$,if $\frac{\Delta [NH_3]}{\Delta t} = 2 \times 10^{-4} \ mol \ L^{-1} s^{-1}$,the value of $\frac{-\Delta [H_2]}{\Delta t}$ would be

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