Question

The scientist "Q" was experimenting to revive a giant called the Mountain. He accidentally dropped a vial of ammonia on the ground. He saw ammonia react with oxygen and produce NO and water.

Let's say the rate of formation of NO is 3.6 x 10^-3 $M{s}^{-1}$

Calculate

• A. 3.6 x 10^-3 $M{s}^{-1}$ and 3.6 x 10^-3 $M{s}^{-1}$
• B. 9 x 10^-3 $M{s}^{-1}$ and 5.4 x 10^-3 $M{s}^{-1}$
• C. 7.2 x 10^-3 $M{s}^{-1}$ and 3.6 x 10^-3 $M{s}^{-1}$
• D. 3.6 x 10^-3 $M{s}^{-1}$ and 5.4 x 10^-3 $M{s}^{-1}$

Answer 1

The correct option is D 3.6 x 10^-3 $M{s}^{-1}$ and 5.4 x 10^-3 $M{s}^{-1}$

The reaction goes as follows:

$4N{H}_3+5O_2\to 4NO+6H_{2}O$

From this we can write the rate equation as follows

$R=-\frac{1}{4}\frac{\Delta \left[{\mathrm{NH}}_3\right]}{\Delta t}=-\frac{1}{5}\frac{\Delta \left[O_2\right]}{\Delta t}=\frac{1}{4}\frac{\Delta \left[\mathrm{NO}\right]}{\Delta t}=\frac{1}{6}\frac{\Delta \left[H_{2}O\right]}{\Delta t}$

Let us try to get the terms we need.

We are given, $\frac{\Delta \left[\mathrm{NO}\right]}{\Delta t}$

which is the rate for formation of NO. We need the rate of disapperance of ammonia.

We know that

$\frac{\Delta \left[\mathrm{NO}\right]}{\Delta t}=-\frac{\Delta \left[{\mathrm{NH}}_3\right]}{\Delta t}$

$\Rightarrow \frac{\Delta \left[{\mathrm{NH}}_3\right]}{\Delta t}=3.6x{10}^{-3}M/s$

Similarly,

$\frac{\Delta \left[H_{2}O\right]}{\Delta t}=\frac{6}{4}\frac{\Delta \left[{\mathrm{NH}}_3\right]}{\Delta t}=\frac{3}{2}\frac{\Delta \left[{\mathrm{NH}}_3\right]}{\Delta t}=5.4x{10}^{-3}M/s$