Question

Chlorine and water react to form hydrogen chloride and oxygen, like this: 2Cl2(g) + 2H2O(g) → 4HCl(g) + O2(g) Also, a chemist finds that at a certain temperature the equilibrium mixture of chlorine, water, hydrogen chloride, and oxygen has the following composition: compound pressure at equilibrium Cl2 34.7atm H2O 33.5atm HCl 12.3atm O2 94.9atm Calculate the value of the equilibrium constant Kp for this reaction. Round your answer to 2 significant digits.

Answer 1

1.6

Step-by-step explanation:

Let's consider the following reaction at equilibrium.

2 Cl₂(g) + 2 H₂O(g) ⇄ 4 HCl(g) + O₂(g)

The pressure equilibrium constant (Kp) is equal to the product of the partial pressures of the products raised to their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients.

`Kp=((pHCl)^(4)* pO_(2) )/((pCl_(2))^(2)* (pH_(2)O)^(2)) =((12.3)^(4)* 94.9 )/((34.7)^(2)* (33.5)^(2)) = 1.61 \approx 1.6`

Answer 2

Answer: The value of
`K_p` for the given chemical reaction is 1.6

Step-by-step explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as
`K_p`

For a general chemical reaction:

`aA+bB\rightarrow cC+dD`

The expression for
`K_p` is written as:

`K_p=(p_(C)^cp_(D)^d)/(p_(A)^ap_(B)^b)`

For the given chemical equation:

`2Cl_2(g)+2H_2O(g)\rightleftharpoons 4HCl(g)+O_2(g)`

The expression for
`K_p` for the following equation is:

`K_p=((p_(HCl))^4* p_(O_2))/((p_(Cl_2))^2(p_(H_2O))^2)`

We are given:

`p_(HCl)=12.3atm\\p_(H_2O)=33.5atm\\p_(Cl_2)=34.7atm\\p_(O_2)=94.9atm`

Putting values in above equation, we get:

`K_p=((12.3)^4* 94.9)/((34.7)^2* (33.5)^2)\\\\K_p=1.6`

Hence, the value of
`K_p` for the given chemical reaction is 1.6