Question #140271
3. A closed system initially containing 1.000 × 10-3 M H2 and 2.000 × 10-3 M I2 at 448℃ is allowed to reach equilibrium. At equilibrium, the HI concentration is 1.87 × 10-3 M. Given that the reaction equation is H2(g) + I2(g) ↔ 2 HI(g), find:
a. Equilibrium concentrations of the two reactants
b. Kc of the reaction taking place.
1
Expert's answer
2020-10-26T14:53:58-0400

H2(g)+I2(g)2HI(g)H_2(g) + I_2(g) ↔ 2 HI(g)

initial concentration

H2=1mMI2=2mMH_2 =1mM\\I_2=2mM

HI=0HI=0


at equilibrium let's assume

H2=(1x)mMI2=(2x)mMHI=2xH_2 =(1-x)mM\\I_2=(2-x)mM\\HI=2x


we are given equilibrium concentration of HI is 1.87mM

2x=1.87x=0.9352x=1.87\\x=0.935


at equilibrium,

H2H_2 =10.935=0.065mM1-0.935=0.065mM

I2=20.935=1.065mMI_2=2-0.935=1.065mM

HI=2×0.935=1.87mMHI=2\times0.935=1.87mM


Kc of the reaction =[HI]2H2I2=1.87×1.871.065×0.065=50.5\frac{[HI]^2}{H_2I_2}=\frac{1.87\times1.87}{1.065\times0.065}=50.5


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