Question

Question: For the reaction

${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Br}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{HBr}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

K_p = 3.5×10⁴ at 1495 K. What is the value of K_p for the following reactions at 1495 K?

a. role="math" localid="1649230142685" $\mathit{H}\mathit{B}\mathit{r}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{B}{\mathit{r}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

b. $\mathbf{2}\mathbf{HBr}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Br}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

c. $\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Br}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{HBr}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Short answer

a. The value of K_p for this reaction at 1495 K is $5.3\times {10}^{ \text{-} 3}$.

b. The value of K_p for this reaction at 1495 K is $2.9\times {10}^{ \text{-} 5}$.

c. The value of K_p for this reaction at 1495 K is 190 .

Step-by-step solution

Write the equilibrium constant expression.

The expression for the K_p for ${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Br}}_2\left(\mathrm{g}\right)\to 2\mathrm{HBr}\left(\mathrm{g}\right)$ is written below:

${\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{{{\mathbf{P}}_{\mathbf{HBr}}}^{\mathbf{2}}}{\left(P_{H_2}\right)\left(P_{{\mathrm{Br}}_2}\right)}$

The given value of K_p for this reaction at 1495 K is $3.5\times {10}^4$

localid="1649233506553" ${\mathrm{K}}_{\mathrm{p}}=3.5\times {10}^4$.

Calculate the Kp value for the given reaction in part a at 1495 K.

The expression for ${\mathrm{K}}_{\mathrm{p}}$for $\mathrm{HBr}\left(\mathrm{g}\right)\to \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{g}\right)$ is written below:

${{\text{K}}^{\text{a}}}_{\text{p}}\text{=}\frac{\left({{\text{P}}_{{\text{H}}_{\text{2}}}}^{\text{1/2}}\right)\left({{\text{P}}_{{\text{Br}}_{\text{2}}}}^{\text{1/2}}\right)}{{\text{P}}_{\text{HBr}}}$

${{\mathbf{K}}^{\mathbf{a}}}_{\mathbf{p}}\mathbf{=}{\left(\frac{1}{K_p}\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}$

Substitute $3.5\times {10}^4$ for ${\mathrm{K}}_{\mathrm{p}}$ into the above expression.

$$\begin{gathered} {{\mathrm{K}}^{\mathrm{a}}}_{\mathrm{p}}=\left(\frac{1}{3.5\times {10}^4}\right) \\ =5.3\times {10}^{-3} \end{gathered}$$

Therefore, the value of ${{\mathrm{K}}^{\mathrm{a}}}_{\mathrm{p}}$ for this reaction at 1495 K is$5.3\times {10}^{ \text{-} 3}$ .

Determine the numerical value of Kp for the given reaction in part b at 1495 K.

The expression for ${{\mathrm{K}}^{\mathrm{b}}}_{\mathrm{p}}$for $2\mathrm{HBr}\left(\mathrm{g}\right)\to {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{Br}}_2\left(\mathrm{g}\right)$ is written below:

${{\text{K}}^{\text{b}}}_{\text{p}}\text{=}\frac{\left({\text{P}}_{{\text{H}}_{\text{2}}}\right)\left({\text{P}}_{{\text{Br}}_{\text{2}}}\right)}{{{\text{P}}_{\text{HBr}}}^{\text{2}}}$

${{\mathbf{K}}^{\mathbf{b}}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{K}}_{\mathbf{p}}}$

Substitute $3.5\times {10}^4$for ${\mathrm{K}}_{\mathrm{p}}$ into the above expression.

$$\begin{gathered} {{\mathrm{K}}^{\mathrm{b}}}_{\mathrm{p}}=\frac{1}{3.5\times {10}^4} \\ =2.9\times {10}^{-5} \end{gathered}$$

Therefore, the value of ${{\mathrm{K}}^{\mathrm{b}}}_{\mathrm{p}}$ for this reaction at 1495 K is $2.9\times {10}^{-5}$.

 Step 4: Find the Kp value for the given reaction in part c at 1495 K.

The expression for ${{\mathrm{K}}^{\mathrm{c}}}_{\mathrm{p}}$for $\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{Br}}_2\left(\mathrm{g}\right)\to \mathrm{HBr}\left(\mathrm{g}\right)$ is written below:

${{\text{K}}^{\text{c}}}_{\text{p}}\text{=}\frac{{\text{P}}_{\text{HBr}}}{\left({{\text{P}}_{{\text{H}}_{\text{2}}}}^{\text{1/2}}\right)\left({{\text{P}}_{{\text{Br}}_{\text{2}}}}^{\text{1/2}}\right)}$

${{\mathbf{K}}^{\mathbf{c}}}_{\mathbf{p}}\mathbf{=}{\left(K_p\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}$

Substitute $3.5\times {10}^4$ for ${\mathrm{K}}_{\mathrm{p}}$into the above expression.

$$\begin{gathered} {{\mathrm{K}}^{\mathrm{c}}}_{\mathrm{p}}=\left(3.5\times {10}^{-4}\right) \\ =190 \end{gathered}$$

Therefore, the value of ${{\mathrm{K}}^{\mathrm{c}}}_{\mathrm{p}}$ for this reaction at 1495 K is 190 .