would the equilibrium constant for reaction be the written as C + A + B? 4. The
ID: 553663 • Letter: W
Question
would the equilibrium constant for reaction be the written as C + A + B? 4. The equilibrium constant for the isomerization of cis-2-butene to trans-2-butene is K-2.07 at 400K. Calculate the standard reaction Gibbs energy (AG n) for the isomerization reaction. (Remember: (AGrn AG n+ RT In K) 5. At 525K, a 20.82 g sample of PCls is placed in a 5.00 L container. At equilibrium, 10.3 g PCl3 have formed. What is the value of K for the reaction? PCIs(g) PCl3(g) + Cl2(g) 6. Suppose a quantity of HI(g) was sealed in a 2.00 L container and brought to equilibrium at 698K. If K-54.3 for the reaction below, and [H11-0.0353M at equilibrium, a) determine the equilibrium concentration values for [H2] and [I2]. b) Also, what is the initial value of [HI]? H2(g) + 12(g) 2H1(g)Explanation / Answer
4) At equillibrium
G = 0
G°rxn + RTlnK = 0
G°rxn = - RTlnK
= - (8.314(J/Kmol) × 400K× ln(2.07)
= - ( 8.314(J/Kmol) × 400K ×2.303 log(2.07))
= - 2.42kJ/mol
5) PCl5(g) <------> PCl3(g) + Cl2(g)
K = [PCl3][Cl2]/[PCl5]
Initial concentration of PCl5
Mass of PCl5 = 20.82g
Molar mass of PCl5 = 208.22g/mol
No of mole of PCl5 = 20.82g/208.22g/ml =0.09999
Volume of Container= 5L
[PCl5] = 0.09999mole/5L =0.0200M
at equillibrium
[ PCl5 ] = 0.0200 - x
[ Cl2 ] = x
[ PCl3 ] = x
Therefore,
K = x^2/(0.0200 - x )
at equillibrium
mass of PCl3 = 10.3g
No of mole of PCl3 =10.3g /137.33g/mol= 0.0750
[PCl3 ] = 0.0750M/5 L = 0.0150M
Therfore,
x = 0.0150
[PCl5] = 0.0200- 0.0150 = 0.0050M
[Cl2] = 0.0150M
[PCl3]= 0.0150M
K =( 0.015)^2/(0.0050)
= 0.045
6) a) H2(g) + I2(g) <-------> 2HI(g)
K = [HI]^2/[H2][I2]= 54.3
At equillibrium
54.3= (0.0353)^2/[H2][I2]
[H2] = x
[ I2 ] = x
x^2=( 0.0353)^2/54.3
= 2.29×10^-5
x = 0.0048
[ H2] = 0.0048M
[ I2 ] = 0.0048M
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