Use data from appendix iib in the textbook to calculate the equilibrium constants at 25 degrees Celsius for each of the following reactions:
a. 2 CO (g) O2(g) <-> 2CO2 (g)
b. 2H2S (g) <-> 2H2 (g) S2 (g)

a) K ≅ 2.0  × 10⁹⁰

b) K = 2.0  × 10⁻²⁶

Explanation:

For the first reaction.

2 CO(g)       +     O₂(g)     ⇄    2CO₂(g)

Equilibrium constant  equation be represented as:

= -RT㏑K

㏑K =             ----------------- equation (1)

Where;

Gas Constant (R) = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹

Temperature (T) = 25°C = (25 + 273 K) = 298 K

Standard Gibbs free energy = ???

To find the    for the above equation; we have:

If we slot in our values back to equation (1); we have:

㏑K  =

㏑K  = 207.6

K =

K = 1.5 × 10⁹⁰

K ≅ 2.0  × 10⁹⁰

Therefore, the equilibrium constants at 25 degrees Celsius for this reaction = 2.0  × 10⁹⁰

b)

For the second reaction.

2H₂S(g)       ⇄    2H₂(g)     +      S₂(g)

Equilibrium constant  equation be represented as:

= -RT㏑K

㏑K =             ----------------- equation (2)

Where;

Gas Constant (R) = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹

Temperature (T) = 25°C = (25 + 273 K) = 298 K

Standard Gibbs free energy = ???

To find the    for the above equation; we have:

Substituting our values, we have :

㏑K  =

㏑K  = -59.13

K =

K = 2.0  × 10⁻²⁶

Therefore, the equilibrium constants at 25 degrees Celsius for this reaction = 2.0  × 10⁻²⁶

it's filtration, but this is an assumption so my apologies if i got this

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year