G @D LA\' 100% Rg. Thu 1:05 p, erPoint Pres Xv . GINsystem ×( e Which one of the

Question

G @D LA' 100% Rg. Thu 1:05 p, erPoint Pres Xv . GINsystem ×( e Which one of the Brooke Wire Practices for Chapter 15 LOs-wi Q- sert Design Layout References Mailings Review View 7. Given the equilibrium constant and initial concentrations, solve for the concentrations at equilibrium 2.L lask, what is the concentration of H2 at A)0.96 B) 0.93 (2) One reaction that occurs in producing steel from iron ore is the reduction of iron(II) oxide the reaction at 1000 K is 0.259. n metal and carbon dioxide. The equilibrium constant, for FcONs) + CO()Fe(s)+ Cog) 0.239 at 1000 K. What is the equilibrium partial pressure of CO at 1000 K if the initial partial pressures are Poo-1.000 strn and PCo.-0.500 atm? A) 1.191 B)0.121 C)0.241 D)0.39 E)0.191 (3) Phosgene is a potent chem Initially, there is 0.50 M COClz in the flask, what are equilibrium concentrations of COC CO, and Cl:? t5 1485 words English (US) NO] = 0.100 M. Find the his 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)

Get Kp expression

Kp = P-CO2 / P-CO

where

Kp = 0.259

find P-CO when...

initially:

PCO = 1

PCO2 = 0.5

in equilbiirum

PCO = 1-x

PCO2 = 0.5+x

substitute in Kp

Kp = P-CO2 / P-CO

0.259 = ( 0.5+x ) /(1-x)

0.259 -0.259 x = 0.5+x

1.259 x =0.259 -0.5

x = (0.259 -0.5)/1.259

x = -0.191

PCO = 1+0.191 = 1.191 atm of CO

PCO2 = 0.5+-0.191 = 0.309

rpoof

Q = (0.309/1.191) = 0.259

which is the same as K, therefore

1.191 atm of CO

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