In the Haber’s process of manufacturing ammunition, N2(g)+3H2(g)→2NH3(g) The rate of appearance of NH3is dNH3dt=2×10−4 mol L−1sec−1 then Rate of disappearance ofH2mol L−1 sec−1Rate of disapperance of N2 mol L−1 sec−1(1)3×10−42×10−4(2)3×10−41×10−4(3)1×10−43×10−4(4)2×10−42×10−4

Rate = +12dNH3dt=−dN2dt=−13dH2dt−dH2dt=32×dNH3dt =32×2×10−4=3×10−4 mol L−1sec−1−dN2dt=12dNH3dt=12×2×10−4 =1×10−4 mol L−1sec−1

In the Haber’s process of manufacturing ammunition, N2(g)+3H2(g)→2NH3(g) The rate of appearance of NH3is dNH3dt=2×10−4 mol L−1sec−1 then Rate of disappearance ofH2mol L−1 sec−1Rate of disapperance of N2 mol L−1 sec−1(1)3×10−42×10−4(2)3×10−41×10−4(3)1×10−43×10−4(4)2×10−42×10−4

Rate = +12dNH3dt=−dN2dt=−13dH2dt−dH2dt=32×dNH3dt =32×2×10−4=3×10−4 mol L−1sec−1−dN2dt=12dNH3dt=12×2×10−4 =1×10−4 mol L−1sec−1

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In the Haber’s process of manufacturing ammunition, $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$ The rate of appearance of $N H_{3}$ is $\frac{d [N H_{3}]}{d t} = 2 \times 10^{- 4}$ mol $L^{- 1} \sec^{- 1}$ then

Rate of disappearance of $H_{2}$
$(m o l L^{- 1} \sec^{- 1})$
Rate of disapperance of $N_{2}$
$(m o l L^{- 1} \sec^{- 1})$
(1) $3 \times 10^{- 4}$ $2 \times 10^{- 4}$
(2) $3 \times 10^{- 4}$ $1 \times 10^{- 4}$
(3) $1 \times 10^{- 4}$ $3 \times 10^{- 4}$
(4) $2 \times 10^{- 4}$ $2 \times 10^{- 4}$

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Detailed Solution

Rate = $+ \frac{1}{2} \frac{d [{NH}_{3}]}{dt} = \frac{- d [N_{2}]}{dt} = - \frac{1}{3} \frac{d [H_{2}]}{dt}$
$\frac{- d [H_{2}]}{dt} = \frac{3}{2} \times \frac{d [{NH}_{3}]}{dt}$
$= \frac{3}{2} \times 2 \times 10^{- 4} = 3 \times 10^{- 4}$ mol $L^{- 1} \sec^{- 1}$
$\frac{- d [N_{2}]}{dt} = \frac{1}{2} \frac{d [{NH}_{3}]}{dt} = \frac{1}{2} \times 2 \times 10^{- 4}$
$= 1 \times 10^{- 4}$ mol $L^{- 1} \sec^{- 1}$

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