Ge(s) + 2H2(g) ----- GeH4 (g) a) given the equilibrium constant for the reaction

Question

Ge(s) + 2H2(g) ----- GeH4 (g)

a) given the equilibrium constant for the reaction is K=2.22 x 10^-4, calculate the concentrations of germane and hydrogen gas at equilibrium if the reaction started with 1.00 atm hydrogen and you have excess Ge Solid Hint: you can use the approximation method for math since Ph2>100 Kp, e.g. the final pressure of the H2 is ~1.00 atm since very little forward reaction takes place.

2H2(g) GeH4(g)

Initial 1.00 0.00

Change

final

answer

b) analysis of the data from above shows that the reaction produces a very low yield of product. Given the high cost of Ge, this is a remedy to the situation. Instead of starting with 1.00 atm hydrogen, start with 100.00 atm hydrogen. Now calculate the amount of germane product at equilibrium. discuss how an excess of the very cheap ingredient H2(g) can cause a dramatic effect on the amount of the expensive product.

2H2(g) GeH4(g)

Initial 100.00 0.00

Change

final

answer

Explanation / Answer

a)since the Ge is in solid state, it wont figure in the equailibrium equation so,

K=[GeH4]/[H2]^2

so let the dissociation be x.so,

K=x/(1-2x)^2

or x/(1-2x)^2=2.22*10^-4

or x=2.218*10^-4 atm

so,

Pressure of H2=1-2*2.18*10^-4

~1 atm

pressure of GeH4=x

=2.18*10^-4 atm

b)since the Ge is in solid state, it wont figure in the equailibrium equation so,

K=[GeH4]/[H2]^2

so let the dissociation be x.so,

K=x/(100-2x)^2

or x/(100-2x)^2=2.22*10^-4

or x=2.04 atm

so,

Pressure of H2=100-2*2.04

=95.92 atm

pressure of GeH4=x

=2.04 atm

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