Question

The differential rate law for the reaction,
$4N{H}_3\left(g\right)+5O_2\left(g\right)\to 4NO\left(g\right)+6H_{2}O(g)$

• A. $-\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=-\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=-\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=-\frac{\text{d}[{\text{H}}_2\text{O}]}{\text{d}\text{t}}$
• B. $\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=-\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=-\frac{1}{6}\frac{\text{d}[{\text{H}}_2\text{O}]}{\text{d}\text{t}}$
• C. $\frac{1}{4}\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=\frac{1}{5}\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=\frac{1}{6}\frac{\text{d}[{\text{H}}_2\text{O}]}{\text{d}\text{t}}$
• D. $-\frac{1}{4}\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=-\frac{1}{5}\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}=\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=\frac{1}{6}\frac{\text{d}[{\text{H}}_2\text{O}]}{\text{d}\text{t}}$

Answer 1

Answer: (D)

The differential rate law for the reaction,
$4\text{N}{\left(\text{H}\right)}_3\left(\text{g}\right)+5{\left(\text{O}\right)}_2\left(\text{g}\right)\to 4\text{N}\text{O}\left(\text{g}\right)+6{\left(\text{H}\right)}_2\text{O}(\text{g})$ is
$\text{R}\text{a}\text{t}\text{e}=-\frac{1}{4}\frac{\text{d}\left[\text{N}{\text{H}}_3\right]}{\text{d}\text{t}}=-\frac{1}{5}\frac{\text{d}\left[{\text{O}}_2\right]}{\text{d}\text{t}}$
$=+\frac{1}{4}\frac{\text{d}\left[\text{N}\text{O}\right]}{\text{d}\text{t}}=+\frac{1}{6}\frac{\text{d}[{\text{H}}_2\text{O}]}{\text{d}\text{t}}$