If the rate of formation of ammonia is 0.345 M/s, what is the rate of disappearance of N₂? $$\frac { 1 } { 2 } \mathrm { H } _ { 2 } ( \mathrm {~g} ) + \frac { 3 } { 2 } \mathrm { H } _ { 2 } ( \mathrm {~g} ) \rightarrow \mathrm { NH } _ { 3 } ( g )$$

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The balanced equation for the formation of ammonia is $ \frac{1}{2} \mathrm{N}_2 + \frac{3}{2} \mathrm{H}_2 \rightarrow \mathrm{NH}_3 $. The stoichiometry shows that 1 mole of $\mathrm{N}_2$ produces 2 moles of $\mathrm{NH}_3$. Therefore, the rate of disappearance of $\mathrm{N}_2$ is half the rate of formation of $\mathrm{NH}_3$, which is $0.345 \, \text{M/s} \times \frac{1}{2} = 0.173 \, \text{M/s}$.

If the Rate of Formation of Ammonia Is 0 A) 0173 M/s B) 0

If the Rate of Formation of Ammonia Is 0 A) 0173 M/s
B) 0