Chapter 10, Problem 10.148QA

Interpretation Introduction

To find:

a. Whether the volume of gases collected can be used to distinguish the pathways of two reactions.

b. If the gas produced during thermal decomposition is $N_{2}O$ or a mixture of $N_{2}and O_2$.

Answer to Problem 10.148QA

Solution:

a. The volume of gases collected can be used to distinguish the pathways of two reactions.

b. The gas produced during the thermal decomposition is $N_{2}O$ gas.

Explanation of Solution

Explanation: 1) Concept:

a. The given two decomposition reactions at two different temperatures produce different number of moles of products. Using this information, we can say that the volume of gases collected can be used to distinguish the pathways of two reactions.

b. As the gaseous products are collected over water, we can get the number of moles of gaseous products $N_2 and O_2$ for the first reaction and $N_{2}O$ for the second reaction. Using the moles of gas, the given temperature, and pressure for both the reaction products, we can calculate the volume. If this volume is the same as that of the given volume, then we can determine whether it is $N_{2}O$ or a mixture of $N_2$ and $O_2$.

2) Formula:  $P\times V=n\times R\times T$      3) Given:

Reactions:

$N{H}_{4}N{O}_3\left(s\right)\stackrel{{300}^{0}C}{\to } N_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)+2H_{2}O\left(g\right)$

$N{H}_{4}N{O}_3\left(s\right)\stackrel{{200-260}^{0}C}{\to } N_{2}O\left(g\right)+2H_{2}O\left(g\right)$

For part b)

i) $mass of N{H}_{4}N{O}_3=0.256 g$

ii) $V=79 mL=0.0079 L$

iii) $T={20}^{0}C$

iv) $P=760 mm Hg$

v) $R=0.08206 atm. L/(mol.K)$

4) Calculations:

a. The decomposition of ammonium nitrate at ${300}^{0}C$, produces one mole of gaseous $N_2$, half mole of gaseous $O_2$, and two moles of gaseous water. So, the total $3.5$ moles of gaseous species produce at ${ 300}^{0}C$ while the same reaction at ${200-260}^{0}C$ produces one mole of gaseous $N_{2}O$, and two moles of gaseous water. So, only $3$ gaseous moles produced by this reaction at $200-{260}^{0}C$. Therefore, we can say that the volume of gases collected can be used to distinguish the pathways of two reactions

b. Finding which gas is produced.

$T={20}^{0}C=20+273 K=293 K$

$P=760 mm Hg=1 atm$

Calculation for moles of $N{H}_{4}N{O}_3$

$0.256 g N{H}_{4}N{O}_3\times \frac{1 mol N{H}_{4}N{O}_3}{80.04 g N{H}_{4}N{O}_3}=0.003198 mol N{H}_{4}N{O}_3$

For first reaction, the number of gaseous moles $=1 mol N_2+\frac{1}{2}mol of O2=1.5$ moles of the mixture of $N_2 and O_2$.

$0.003198 mol N{H}_{4}N{O}_3\times \frac{1.5 mol mixture of N_2 and O_2}{1 mol N{H}_{4}N{O}_3}$

$=0.004798 mol mixture of N_2 and O_2=n$

Therefore, volume calculation for first reaction is

$1 atm\times V=0.004798 mol\times 0.08206 \frac{atm.L}{\left(mol.K\right)}\times 293 K$

$V=\frac{\left[0.004798 mol\times 0.08206 \frac{atm.L}{\left(mol.K\right)}\times 293 K\right]}{1 atm}$

$V=0.115 L$

For second reaction, $n=1 mol of N_{2}O$,

$0.003198 mol N{H}_{4}N{O}_3\times \frac{1 mol N_{2}O}{1 mol N{H}_{4}N{O}_3}=0.003198 mol N_{2}O=n$

Therefore, volume calculation for second reaction is

$1 atm\times V=0.003198 mol\times 0.08206 \frac{atm.L}{\left(mol.K\right)}\times 293 K$

$V=\frac{\left[0.0032 mol\times 0.08206 \frac{atm.L}{\left(mol.K\right)}\times 293 K\right]}{1 atm}$

$V=0.0769 L$

Second reaction’s volume is nearly the same as the given volume. Therefore, the gas produced is $N_{2}O$.

Conclusion:

a.  The number of the gaseous moles of the product are different for both the reaction. So we can say that their volumes collected can be used to distinguish their pathways.

b. The volume of the product produced by the given amount of ammonium nitrate can help to get which gas is produced.