Question

Consider the following reactions:

${\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathit{H}\mathit{I}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{I}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}\mathit{H}\mathit{I}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

List two property differences between these two reactions that relate to equilibrium.

Short answer

1. The equilibrium constant expression for Reaction (1) differs from Reaction (2).
   K is equal to K_p for Reaction (1) because $∆n_1=0$, K and is not equal to K_p for Reaction (2) due to the non-zero value of $∆n_2$.
2. The change in container’s volume has no influence on reaction (1) whereas the change in container’s volue has influence on the reaction (2).

Step-by-step solution

Compare the equilibrium constant between two reactions.

The given balanced chemical equation is shown below:

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ …(1)

The corresponding expression for the equilibrium constant, K, is written below:

$K=\frac{{\left[HI\right]}^2}{\left[H_2\right]\left[I_2\right]}$ …(2)

Another balanced chemical equation is given below:

$H_2\left(g\right)+I_2\left(s\right)\rightleftharpoons 2HI\left(g\right)$ …(3)

The equilibrium constant, K expression of the above reaction is written below:

$K=\frac{{\left[HI\right]}^2}{\left[H_2\right]}$ …(4)

The concentration of iodine is ignored because it is in a solid state.

The equilibrium constant expression for Reaction (1) is different from Reaction (2).

Determine the change in mole for both reactions.

The below expression that describes the relationship between K_p and K is:

$K_p=K{\left(RT\right)}^{∆n}$ …(5)

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.

$∆n$is the numerical difference between the coefficients of gaseous products and the coefficients of gaseous reactants.

The value of $∆n$for Reaction (1) can be calculated as follows:

$\begin{array}{rcl}∆n_1& =& 2-(1+)\\ & =& 0\end{array}$

Substitute 0 for $∆n$into Equation (5).

$\begin{array}{rcl}{K}_p& =& K{\left(RT\right)}^0=K\end{array}$

The value of $∆n_2$for Reaction (2) can be calculated as follows:

$\begin{array}{rcl}∆n_2& =& 2-\left(1\right)\\ & =& 1\end{array}$

K is equal to K_pfor Reaction (1)because $∆n_1=0$ , whereas K is not equal to K_p for Reaction (2) due to the non-zero value of $∆n_2$.

The change in container’s volume has no influence on reaction (1) whereas the change in container’s volue has influence on the reaction (2) and the equilibrium favors the foration of products on increasing the volume of the container.