For the following reaction of gases, Kc =2.00 at 1000degC 2COF2 <--> CO2 + CF4 I

Question

For the following reaction of gases, Kc =2.00 at 1000degC 2COF2 <--> CO2 + CF4 If a 5.00L mixture contains 0.145 mol COF2, 0.262 mol CO2, and 0.074 mol CF4 at 1000 degC

A. Find Q for this reaction mixture.

B. Predict the direction in which a net change will occur.

C. How many moles of each gas will be present at equilibrium?

D. What is the concentration of each species at equilibrium?

I have the answers but can someone please show the work for Part C & D? Thanks!

A. Q = 0.9221403092
B. Right
C. 0.1121837011, 0.2784081495, 0.0904081495
D.0.0224367402, 0.0556816299, 0.0180816299

Explanation / Answer

Kc = [CO2] [CF4] / [COF2]^2

when the numerator has molarity squared , & the bottom has molarity squared,,,
it makes no difference whether you use molarities or whether you use moles in the [ ],
the units cancel out

Q = [0.262] [0.074] / [0.145]^2

Q = 0.922 which is too small, compared to the Kc of 2.00

i) will the mixture be at equilibrium?..... no
====================================

(ii) if the gases are not at equilibrium, in what direction will a net change occur?
it will shift to the right producing more -->CO2(g) + CF4(g).... inorder to increase the numerator 7 decrease the denominator ,,, until the ratio is = Kc
======================================...

(iii) determine the number of moles of each gas at equilibrium.
2 COF2(g) --> CO2(g) & CF4(g)
0.145 - 2X --> (0.262 +X) & (0.074 +X)

Kc = [CO2] [CF4] / [COF2]^2

2.00 = [0.262 +X] [.074 +X] / [0.145 - 2X ]^2

(2.00) [0.145 - 2X ]^2 = [0.262 +X] [.074 +X]

(2.00) [0.021 +4X2 -0.58X ] = 0.0194 + X2 + 0.336X

0.042 +8X2 - 1.16X = 0.0194 + X2 + 0.336X

7X2 - 1.50X & 0.0226 = 0

X = 0.0163

the number of moles of each gas at equilibrium.
COF2(g) = 0.145 - 2X = 0.145 - 0.032 = 0.113 moles
CO2(g) = (0.262 +X) = 0.262 + 0.0163 = 0.278 moles
CF4(g) = (0.074 +X) = 0.074 + 0.0163 = 0.0903 moles

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