Question

Given the equilibrium constants for the following two reactions at a 298K:NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600Calculate the value for the equilibrium constant, Kc, for the reaction:CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Answer 1

To calculate the equilibrium constant Kc for the reaction CO2(g) + H2(g) ⇌ CO(g) + H2O(g), multiply the equilibrium constants for the given reactions (40 and 600) to obtain 24000.

Step-by-step explanation:

To calculate the equilibrium constant, Kc, for the reaction CO2(g) + H2(g) ⇌ CO(g) + H2O(g), we can use the equilibrium constants for the given reactions. First, we need to combine the given reactions to obtain the desired reaction. We can reverse the second reaction and multiply the two reactions together, cancelling out the common species Ni and CO2:

NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) (1)

CO2(g) + Ni(s) ⇌ CO(g) + NiO(s) (reversed and multiplied by reaction 1)

By multiplying the two reactions together, we cancel out Ni and CO2:

H2(g) + CO2(g) ⇌ CO(g) + H2O(g)

Now, we can calculate the equilibrium constant for the desired reaction by multiplying the equilibrium constants for the given reactions:

Kc = Kc1 * Kc2 = 40 * 600 = 24000

Answer 2

The value is
`K_C = (40)/(600)`

Step-by-step explanation:

From the question

The equation given is

NiO(s) + H2(g) ⇌ Ni(s) + H2O(g) Kc=40 (1)

NiO(s) +CO(g) ⇌ Ni(s) +CO2(g) Kc=600 (2)

Generally the reverse of the second equation as shown below

Ni(s) +CO2(g) ⇌ NiO(s) + CO(g) (3)

The equilibrium constant becomes
`K_c ' = (1)/(600)`

Now adding 1 and 3 we obtain

NiO(s) + H2(g)+ Ni(s) +CO2(g) ⇌ Ni(s) + H2O(g)+NiO(s) + CO(g)

CO2(g) + H2(g) ⇌ CO(g) + H2O(g)

Hence the equilibrium constant for the resulting equation is mathematically evaluated as

`K_C = K_c' * K_c`

=>
`K_C = (1)/(600) * 40`

=>
`K_C = (40)/(600)`