Question

Calculate ΔG° for the following reaction from the equilibrium constant at the temperature given. HF(aq)+H2O(l)⟶H3O+(aq)+F−(aq),T=25°C,K=7.2×10−4

Answer 1

Answer:

17.93 kJ

Explanation:

The relation between standard Gibbs energy and equilibrium reaction is shown below as:

$\Delta{G^0} =-RT \ln K$

R is Gas constant having value = 0.008314 kJ / K mol  

Given For the reaction:-

$HF+H_2O\rightarrow H_3O^++F^-$

temperature, T = $25^oC=[25+273]K=298K$

$K=7.2\times 10^{-4}$

Applying in the above equation as:-

$\Delta{G^0} =-0.008314\times 298 \ln(7.2\times 10^{-4})\ kJ=17.93\ kJ$