Question

For the reaction $2NH_{3} \rightarrow N_{2} + 3H_{2}$if $- \frac { \mathrm { d } \left[ \mathrm { NH } _ { 3 } \right] } { \mathrm { dt } } =$ $\frac{d\lbrack N_{2}\rbrack}{dt} = k_{2}\lbrack NH_{3}\rbrack,\frac{d\lbrack H_{2}\rbrack}{dt} = k_{3}\lbrack NH_{3}\rbrack$

Then the relation between $k_{1},k_{2}$ and $k_{3}$ is

• A. $k_{1} = k_{2} = k_{3}$
• B. $k_{1} = 3k_{2} = 2k_{3}$
• C. $1.5k_{1} = 3k_{2} = k_{3}$
• D. $2k_{1} = k_{2} = 3k_{3}$

Answer 1

Answer: (C)

$1.5k_{1} = 3k_{2} = k_{3}$

Answer 2

$2NH_{3}\overset{\quad\quad}{\rightarrow}N_{2} + 3H_{2}$

Rate = $- \frac{1}{2}\frac{d\lbrack NH_{3}\rbrack}{dt} = \frac{d\lbrack N_{2}\rbrack}{dt} = \frac{1}{3}\frac{d}{dt}$

$\frac{1}{2}k_{1}\lbrack NH_{3}\rbrack = k_{2}\lbrack NH_{3}\rbrack = \frac{1}{3}k_{3}\lbrack NH_{3}\rbrack$

$1.5k_{1} = 3k_{2} = k_{3}$