Question

The decomposition $N{H_3}$ gas on a heated surface gave the following results:

Initial Pressure(mm)	$65$	$105$	y	$185$
Half-life(sec)	$290$	x	$670$	$820$

Calculate approximately the values of x and y.
x= $410$ sec, y= $115$ mm
x= $467$ sec, y= $150$ mm
x= $490$ sec, y= $120$ mm
x= $430$ sec, y= $105$ mm

Answer 1

When ammonia is heated on the surface of catalyst tungsten, it decomposes into nitrogen gas and hydrogen gas. This reaction is a zero-order reaction because the rate of reaction does not depend on the concentration of reacting species. Use the formula which shows the relation of rate constant for zero-order reaction with initial pressure and half-life-
$\Rightarrow {\text{k = }}\dfrac{{\left[ {{{\text{P}}_{\text{0}}}} \right]}}{{{\text{2}}{{\text{t}}_{{\text{1/2}}}}}}$ to find the value of k. Then use the same formula to find the value of x and y.

Complete Step-by-Step Solution:
We know that the half life for nth order is inversely proportional to initial pressure raised to the power$\left( {n - 1} \right)$. It is written as-
$\Rightarrow {{\text{t}}_{{\text{1/2}}}} \propto \dfrac{{\text{1}}}{{{{\text{P}}^{{\text{n - 1}}}}}}$ Where n is the order of reaction, ${{\text{t}}_{1/2}}$ is half life and P is initial pressure.
So we can write it as-
$\Rightarrow {{\text{t}}_{{\text{1/2}}}} \propto {{\text{P}}^{{\text{1 - n}}}}$ --- (i)
Then according to given data we can write,
$\Rightarrow \dfrac{{290}}{{820}} = \dfrac{{65}}{{185}}$
On dividing we see that both are equal so we can say that half life is equal to initial pressure according to data, then from eq. (i) we can write,
$\Rightarrow {\text{P = }}{{\text{P}}^{{\text{1 - n}}}}$
Since the base on both sides is same so the powers of bases should also be equal so we get,
$\Rightarrow {\text{1 = 1 - n}}$
On solving we get,
$\Rightarrow n = 0$
This means that decomposition of ammonia on the surface of tungsten is zero order reaction.
Now we have to find the value of x and y in the data.
We know that the relation of the rate constant for zero-order reaction with initial pressure and the half-life is given as-
$\Rightarrow {\text{k = }}\dfrac{{\left[ {{{\text{P}}_{\text{0}}}} \right]}}{{{\text{2}}{{\text{t}}_{{\text{1/2}}}}}}$ -- (ii)
Then on putting the values from the given data we can get,
$\Rightarrow {\text{k}} = \dfrac{{65}}{{2 \times 290}}{\text{ or }}\dfrac{{185}}{{2 \times 820}}$
On solving any of the above function we get,
$\Rightarrow {\text{k = 0}}{\text{.112 mol}}{{\text{L}}^{ - 1}}{{\text{t}}^{ - 1}}$
Now we know the of k so we can also write the eq. (i) as-
$\Rightarrow {{\text{t}}_{{\text{1/2}}}}{\text{ = }}\dfrac{{\left[ {{{\text{P}}_{\text{0}}}} \right]}}{{{\text{2k}}}}$
So on putting ${{\text{t}}_{{\text{1/2}}}} = x$ and P=$105$ and also putting the given vale of k we get,
$\Rightarrow x = \dfrac{{105}}{{2 \times 0.112}}$
On solving we get,
$\Rightarrow$ x=$467{\text{ sec}}$
Now for y, we can write the formula as-
$\Rightarrow \left[ {{{\text{P}}_{\text{0}}}} \right]{\text{ = 2k}} \times {{\text{t}}_{{\text{1/2}}}}$
On putting the given values we get,
$\Rightarrow y = 2 \times 0.112 \times 670$
On multiplication we get,
$\Rightarrow y = 150{\text{ mm}}$

Answer-Hence the correct answer is B.

Note: The characteristics of zero-order reactions are-
1.The rate of reaction of these reactions does not change with change in concentration.
2.For the zero-order reaction, the graph plotted between the concentration and rate is a straight line parallel to the concentration axis.
3.The zero-order reaction always occurs in the presence of catalysts like certain enzyme-catalyzed reactions and photochemical reactions.
4.The half-life period for the zero-order reaction is directly proportional to the initial concentration and inversely proportional to the rate constant.