Question

2. Nitrogen monoxide (NO) and nitrous oxide (N20) gases can both undergo decomposition to their elemental constituents at high temperature. a. Write balanced chemical equations for each of these decomposition reactions. b. Consider a mixture of NO and N20 that is placed into a seal reaction chamber at 300 K with an initial total pressure of 380 torr. After complete decomposition of both gases, the total pressure in the chamber is 1900 torr at 1400 K. What is the percent of nitrous oxide in the original mixture?

Answer 1

a.

NO decomposes to molecular N₂ and molecular O₂ as follows:

2NO(g) + N2(g) + O2(g)

N₂O decomposes to molecular N₂ and molecular O₂ as follows:

21.09) + 2N2(g) + O2(g)

Note that both the equations above are balanced equations.

b.

Now, initial pressure at 300 K is about 380 Torr when we have only NO and N₂O in the reaction chamber.

Hence, using the ideal gas law, we can write the following equation for the total number of moles of NO and N₂O

PV=nRT n= RT 380 Torr XV n= RX 300 K 267

Let the number of moles of NO initially be x moles, and number of moles of N₂O initially be y moles.

Hence,

n = x + y = 1.267-

Where V is the volume of the reaction chamber.

R is the gas constant.

Note that both V and R are constant in our case.

For complete decomposition of x moles of NO, we can write the following ICF table

	2NO	$\rightarrow N_2$	$+O_2$
Initial, mol	x	0	0
Change, mol	-x	$+\frac{x}{2}$	$+\frac{x}{2}$
Final, mol	0	$\frac{x}{2}$	$\frac{x}{2}$

Hence, total number of moles of gas formed from complete decomposition of x moles of NO is

.

+

Hence, x moles of NO results in total x moles of gases being produced.

For complete decomposition of y moles of N₂O, we can write the following ICF table

	2N2O	+ 2N	$+O_2$
Initial, mol	y	0	0
Change, mol	-y		$+\frac{y}{2}$
Final, mol	0	y	$\frac{y}{2}$

Hence, total number of moles of gas formed from complete decomposition of y moles of N₂O is

.

+

Hence, y moles of N₂O results in total of 3y/2 moles of gases produced.

It is given that the total pressure in the chamber is 1900 torr at 1400 K after complete decomposition.

Hence, using the ideal gas law, we can write

PV = nRI RT 3y 1900 torr XV 2 - Rx 1400 K 1.357

Hence, we have two equations for x and y, they are

x+y= 1.267......(1)

7 + = 1.357 ....(2

Hence, using substitution of in equation (2), we can write

x = 1.267 - y

= 1.267– y + = 1.357 = = (1.357 – 1.267) = 0.090 0810 = 5060'0 x 7=13

Note that y is the number of moles of N₂O in the original mixture.

Hence, percentage N₂O in the original mixture can be calculated as follows:

1 0.180 x 100 = 100 1.267 x 100 – 14.3%

Hence, the answer is 14.3 % of all gas molecules in the original mixture is nitrous oxide.

Note: this % is not weight percentage but rather the mole % of N₂O in the original mixture.