POSTLABORATORY ASSIGNMENT 1. Calculate the average and standard deviation values

Question

POSTLABORATORY ASSIGNMENT 1. Calculate the average and standard deviation values for Kc 2. Hydrogen and carbon dioxide gas react at high temperature to give water and carbon monoxide H2(g) + CO2(g)=H2O(g) + CO(g) a) Measurements at 9860C show that there is 0.11 mole of each of CO and H20 vapor and 0.087 mole each of H2 and CO2 at equilibrium in a 2.0 L container. Calculate the equilibrium constant for the reaction at 986°c. (show your calculations) b) Suppose 0.050 mole each of H2 and CO2 are placed in a 2.0 L container. When equilibrium is achieved at 986°C what amounts of CO and H20, in moles will be present? (Use Kc from part a) (show your calculations)

Explanation / Answer

Q2

a)

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

Keq = [H2O][CO]/([H2][CO2])

M = mol/V = mol /2

K =(0.11/2)(0.11/2)/((0.087/2)(0.087/2))

K = 1.5986

b)

if

[H2] = 0.05/2 = 0.025

[CO] = 0.05/2 = 0.025

find CO and H2O in equilbirium

1.5986

Keq = [H2O][CO]/([H2][CO2])

[H2O] = [CO] = x

[H2] = [CO2]= 0.025 - x

1.5986 = (x*x)/(0.025-x)(0.025-x)

sqrt(1.5986 ) = x / (0.025-x)

1.264*(0.025 - x) = x

1.264*(0.025) - 1.264x = x

(1+1.264)x = 1.264*(0.025)

x = 1.264*(0.025) / 2.264

x = 0.01395

[H2] = [CO2]= 0.025 - 0.01395 = 0.01105 M

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