Question

At 1100 K, K_p= 0.25 for the reaction

$\mathbf{2}\mathit{S}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{2}\mathit{S}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

What is the value of K at this temperature?

Short answer

The value of equilibrium constant, K at 1100 K is 23.

Step-by-step solution

The relation between Kp   and  Kc

The expression that describes the relationship between K_p and K is,

${\text{K}}_{\text{p}}\text{=K}{\left(\text{RT}\right)}^{\text{∆n}}$ …(1)

Here,

K_p is the equilibrium constant in terms of partial pressures.

K is the equilibrium constant in terms of concentrations.

T is the temperature.

R is the universal gas constant.

$∆\mathrm{n}$is the numerical difference between the coefficients of gaseous products and the coefficients of gaseous reactants.

 Step 2: Calculation

The value of for the given reaction can be calculated as follows:

$\begin{array}{rcl}& & ∆\mathrm{n}\text{= 2} \text{--} \left(\text{2 + 1}\right)\\ & & \text{=} \text{-} \text{1}\\ & & \end{array}$

Substitute $\text{-} \text{1}$for $∆\mathrm{n}$, 0.25 for ${\text{K}}_{\text{p}}$, $\text{0.08206} \text{L} \text{atm/mol} \text{K}$for R, and 1100 K for T into Equation (1).

$\text{0.25 = K}{\left(\text{0.08206} \text{L} \text{atm/mol} \text{K×1100} \text{K}\right)}^{\text{-1}}$

Rearrange to find K.

$\begin{array}{rcl}& & \text{K = 0.25×0.08206} \text{L} \text{atm/mol} \text{K×1100} \text{K}\\ & & \text{= 22.5665}\\ & & \text{= 23}\\ & & \end{array}$

Hence, the value of equilibrium constant, K at 1100 K is 23.