Question

Given that $\Delta G(HI:g) \equiv + 1.7kJ$ What is the equilibrium constant at ${25^0}C$ for $2HI(g) \rightleftarrows {H_{2\left( g \right)}} + {I_{2\left( g \right)}}$ ?
A. $24.0$
B. $3.9$
C. $2.0$
D. $0.5$

Answer 1

We have to remember that an equilibrium constant is expressed by $K$ , the number which expresses the relationship between the products and reactants present at equilibrium in a reversible chemical reaction.
Here, we use change in free energy for a reaction at any conditions other than standard can be calculated as,
$\Delta {G^0} = \Delta G + RT\ln K$ , for standard conditions
$T = {25^0}C$ ,
R and K are constants.

Complete step by step answer:
Given, $\Delta G(HI:g) \equiv + 1.7kJ$
$T = {25^0}C$ ,
$2HI(g) \rightleftarrows {H_{2\left( g \right)}} + {I_{2\left( g \right)}}$ .
Rearranging the above change in free energy equation, we get
$\Delta G^\circ = - 2.303 \times RT{\log _{10}}{K_{\left( {eq} \right)}}$
Where,
$\Delta G =$Change in Gibbs free energy
$R$ is the gas Constant, value is $8.314J{K^{ - 1}}mo{l^{ - 1}}$ or $0.008314kJmo{l^{ - 1}}{K^{ - 1}}$ .
$T$is the temperature in kelvin.
$K$is the equilibrium Constant.
In equilibrium, the temperature in Kelvin, but in question, given in Celsius form, so, degree Celsius is converted to kelvin,
$K = T + 273$
$K = 25 + 273$
$K = 298$
At equilibrium,
$R = 8.314$.
$\Delta {G^0} = - 2.303 \times RT{\log _{10}}K$
On substituting the known values we get,
$\Rightarrow 1.73 \times {10^3} = - 2.303 \times 8.314 \times {\log _{10}}({K_{eq}}) \times 298$
$\Rightarrow 1700 = - 2.303 \times 8.314 \times {\log _{10}}({K_{eq}}) \times 298$
On simplification we get,
$1700 = - 5705 \times {\log _{10}}({K_{eq}})$
Rearranging the above equation,
$\dfrac{{1700}}{{ - 5705}} = {\log _{10}}({K_{eq}})$
On simplification we get,
$- 0.297 = {\log _{10}}\left( {{K_{eq}}} \right)$
Rearranging the above equation we get,
$\Rightarrow {K_{eq}} = {10^{( - 0.297)}}$
$\Rightarrow {K_{eq}} = 0.5$

So, the correct answer is Option D.

Additional information:
In the above calculation we see Gibbs free energy, called “Available energy”-Gibbs free energy, was developed by Josiah Willard Gibbs, an American scientist. It is defined by Josiah Willard Gibbs, developer, “a certain substance in a given initial state give the greatest amount of mechanical work which can be obtained from a given quantity, allowing heat to pass or from external bodies or without increasing its total volume, except such as at the close of the process are left in their initial condition.

Note: In $\Delta G =$ Change in Gibbs free energy, the mean “available in the form of useful work”. $\Delta {G^0}$ indicates all reactants and products are in their standard states. And also for standard conditions $\Delta G = 0$ , for a system at equilibrium. Not confusing with Helmholtz free energy, which is represented by the symbol$- {A^0}$ , which gives maximum work obtainable from a system. But $- \Delta G$ , which gives maximum useful work obtainable from the system.