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à Secure https://session.masteringchemistry Assignment 3 (Chapter 14) Problem 14

ID: 1075346 • Letter: #

Question

à Secure https://session.masteringchemistry Assignment 3 (Chapter 14) Problem 14.92 ID-95478478 ) 7of9 Constants 1 Periodic Table PartA The air polutant NO is produced in automobile engines from the high-temperature reaction Calculate the equilibrium concentration of [N: if the initial concentrations are 2.25 M N2 and 0.57 M Op- (This N2/O2 concentration ratio is the ratio found in air.) Express your answer to three significant figures and include the appropriate units. (g)Olg)-2N0(9) 20. K,=1.7 ×10-3 at N-Value Units Submit PartB Calculate the equlibrium concentration of (O Express your answer to two significant figures and include the appropriate units. [O2] .1 Value Units | Submit Request Answer Part C

Explanation / Answer

Given that

N2(g)+O2(g)<-> 2NO(g) Kc= 1.7*10^-3 at 2300

let x amount of product is formed

..N2..........O2..........2NO
I...2.25.........0.57..........0
C...-x.............-x...........+2x
E..2.25-x........0.57-x.......2x

Kc = 1.7x10^-3 = [NO]^2 / [N2][O2]
1.7x10^-3 = (2x)^2 / (2.25-x)(0.57-x)

1.7x10^-3 = 4x^2 / 1.28 -2.25 x-0.57 x +x^2


1.7x10^-3 = 4x^2 / (x^2 - 2.82x + 1.28)


1.7x10^-3x^2 - 0.0048x + 0.002 = 4x^2
3.998 x^2 + 4.8x10^-3x – 2.2x10^-3 = 0

X= 0.0229 M

[NO] = 2*0.0229 M

= 0.0458 M

[N2] =2.25-x= 2.25- 0.0229 =2.2271

[O2]= 0.57-x=0.57-0.0229 M

= 0.5471 M

Kc = 1.7x10^-3 = [NO]^2 / [N2][O2]

=0.04582/ 2.2271* 0.5471

= 1.7*10^-3