Question

Consider the following reaction at 298 K: 2H2S(g)+SO2(g)→3S(s, rhombic)+2H2O(g),ΔG∘rxn=−102 kJ Calculate ΔGrxn under these conditions: PH2SPSO2PH2O===2.00 atm1.50 atm0.0100 atm

Answer 1

Answer: The $\Delta G$ for the reaction is 74.732 kJ/mol

Explanation:

For the given chemical reaction:

$2H_2S(g)+SO_2(g)\rightleftharpoons 3S\text{(s, rhombic)}+2H_2O(g)$

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{H_2O})^2\times (p_{S\text{s,rhombic}})^3}{(p_{H_2S})^2\times p_{SO_2}}$

We are given:

$p_{H_2O}=0.0100atm\\p_{H_2S}=2.00atm\\p_{SO_2}=1.50atm\\p_{S\text{s, rhombic}}=1atm$

Partial pressure of solids are taken as 1.  

Putting values in above equation, we get:

$K_p=\frac{(0.0100)^2}{(2.00)^2\times 1.50}\\\\K_p=1.66\times 10^{-5}$

To calculate the Gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs free energy of the reaction = ?

$\Delta G^o$ = Standard Gibbs' free energy change of the reaction = 102 kJ = 102000 J     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.314J/K mol$

T = Temperature = 298 K

$K_p$ = equilibrium constant in terms of partial pressure = $1.66\times 10^{-5}$

Putting values in above equation, we get:

$\Delta G=102000J+(8.314J/K.mol\times 298K\times \ln(1.66\times 10^{-5}))\\\\\Delta G=74731.6J/mol=74.732kJ/mol$

Hence, the $\Delta G$ for the reaction is 74.732 kJ/mol