Question

Consider the reaction
$H_2\left(g\right)+B{r}_2\left(g\right)\iff 2HBr\left(g\right)$
where$\Delta H^o=-103.8kJ$ . In a particular experiment,1.00 atm of $H_2\left(g\right)$and 1.00 atm of $B{r}_2\left(g\right)$were mixed ina 1.00-L flask at 25°C and allowed to reach equilibrium.Then the molecules of$H_2$ , were counted by using a very sensitive technique, and $1.10\times {10}^{13}$molecules were found. For this reaction, calculate the values of K,$\Delta G^o$, and$\Delta S^o$ .

Short answer

$K=2.0\times {10}^{19}$

$\Delta G°=-1.10\times {10}^{5}J/mol$

$∆S^{°}=+20.805J/molK$

Step-by-step solution

Introduction to the Concept

Entropy is a characteristic of a system that quantifies the unpredictability or degree of chaos. It’s a state function, after all. With an increase in the universe’s entropy, spontaneous change occurs.

The mathematical equation for the standard entropy value at room temperature is given as follows:

In the balanced chemical equation, n and p indicate the coefficients of reactants and products.

$\Delta G^o=\Delta H^o-T\Delta S$

$\Delta G^o=-RT\times ln\times K$

Where

R = Universal gas constant ($R=8.314J/mol-K$)

T = Temperature

Determination ΔGoR at 25oC and 1 atm

The given reaction is:

$H_2\left(g\right)+B{r}_2\left(g\right)\iff 2HBr\left(g\right)$

1.00L is the volume of the flask in which the reaction is taking place.

At a pressure of 1 atm, equal moles of $H_2\left(g\right)$and $B{r}_2\left(g\right)$ are present.

$PV=nRT$is the ideal gas equation.

The number of moles of each $H_2\left(g\right)$and $B{r}_2\left(g\right)$ present in the flask at the start of the experiment:

$n=\frac{PV}{RT}$

$n=\frac{1.00atm\times 1.00L}{0.08206L\times atm/mol\times K\times 298K}$

$n=0.041\text{mol}$

The reaction’s equilibrium is shown below:

The number of moles at the equilibrium of $H_2\left(g\right)=1.10\times {10}^{13}molecules$

Determination of H2g  at equilibrium

$=\frac{1.10\times {10}^{13}\text{molecules}\times \text{1 mole}\times {\text{H}}_2\left(g\right)}{6.023\times {10}^{23}\text{molecules}}$

$n=1.83\times {10}^{-11}{\text{moles ofH}}_2\left(g\right)$

Therefore,

$x=0.041\text{moles}$, since $1.83\times {10}^{-11}<<<<0.041$

Here, the equilibrium constant:

$K=\frac{{\left[HBr\right]}^2}{\left[H_2\right]\times \left[B{r}_2\right]}$

$=\frac{{\left[2x\right]}^2}{\left[0.041-x\right]\times \left[0.041-x\right]}$

$K=2.0\times {10}^{19}$

The $\Delta G^o$and the value of its equilibrium constant at 298 K:

$\Delta G^o=-R\times T\times \ln \times K$

$∆G^o=(-8.314J/mol.K)\times (298K)\times \ln (2.0\times {10}^{19})$

$=-1.10\times {10}^{5}J/mol$

Determination of ΔGo at equilibrium

$\Delta G^o=\Delta H^o-T\Delta S$

$∆H^o=-103.8kJ/mol$

$=-103800$

$∆G^o=-1.10\times {10}^{5}J/mol$

Let’s substitute all values

$-1.10\times {10}^{5}J/mol=-103800J/mol-298(∆S^o)$

$298(∆S^o)=110000J/mol-103800J/mol$

$\left(\Delta S°\right)=\frac{110000J/mol-103800J/mol}{298K}$

$\Delta S°=+20.805J/molK$

Therefore the $\Delta S^o$is$20.805J/molK$ .