ALEKS-Learn-R, × My Blackboard Co × YouTube Google Sc × Boston Celtics IBi (1 Se

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ALEKS-Learn-R, × My Blackboard Co × YouTube Google Sc × Boston Celtics IBi (1 Secure https/lwww.awh.aleks.com/alekscgi/x/lsl.exerto u-1gNsikr7j8P3)4-UjTgkogHgD-hT/LCmFKBGScXhEWOIL Tr S1J O ADVANCED MATERIAL Calculating equilibrium composition from an equilibrium constant Suppose a 250. mL. flask is filled with 1.1 mol of 12 and 0.50 mol of HI. The following reaction becomes possible: H2(g)+128)2HI(g) The equilibrium constant K for this reaction is 0.174 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to two decimal places Check MacBook Air

Explanation / Answer

Concentrations = moles/volume, volume has to be in liters, 250ml= 250/1000L=0.25L

Concentrations : I2= 1.1/0.25= 4.4M and HI= 0.5/0.25= 2M

The reaction is H2(g)+ I2(g) <----->2HI(g)

Q= reaction coefficient = [HI]2/ [I2][H2] =2/(4.4*0)= infinity >K. Hence Q has to be reduced so as to reach equilibbrium value. This is possible only if the HI is decompsed

Let x= moles of H2 formed at equilibrium,

At Equilibrium [H2]= x and [I2] =4.4+x and [HI]= 2-2x

K= (2-2x)2/{(x*(4.4+x)}=0.174

When solved for x using excel, , x= 0.629

Concentration of H2 at equilibrium = 0.629M

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