100% sa, Thu 1:38 PM PowerPoint Pres ×·GNsystem ×Ye which one of thx) Brook Prac

Question

100% sa, Thu 1:38 PM PowerPoint Pres ×·GNsystem ×Ye which one of thx) Brook Practices for Chapter 15 LOS-w. a Insert Design Layout References Mailings Review View Times New Ro-.-112 , A-Ar A: .abe x, x' A. " A. Styles CO, and Cla? COCk]-0.37 C) (4) The research and development unit of a chemical company is studying the reaction of CH4 and H:S, two components of natural gas: In one experiment, 1.00 mol of CHa, 1.00 mol of CS, 2.00 mol of H:S, and 2.00 mol of H are mixed in a 250-mL vessel at 960 C. At this temperature, K-0.036. If (CH]-5.56 M at equilibrium, what are the equilibrium concentrations of H:? A)556 M B) 1.76 M C) 6.44 M D) 244 M of 5 1485 Wards English (US [NO] = 0.100 M. Find the this temperature.

Explanation / Answer

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)

K = [CS2][H2]^4 /([CH4][H2S]^2)

calcualte molarities via, M = mol/V

[CH4] = 1/0.25 = 4

[CS2] = 1/0.25 = 4

[H2S] = 2/0.25 = 8

[H2] = 2/0.25 = 8

Since CH4 increases, then the reaction goes reverse so..

in equilbirum

[CH4] = 4+x

[CS2] = 4+2x

[H2S] = 8-x

[H2] = 8-4x

and we know

[CH4] = 4+x = 5.56; x = 5.56-4 = 1.56

then

[H2] = 8-4x = 8-4*1.56 = 1.76 M

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