Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Calculate the standard Gibbs free energy change for the following reaction at <img alt="\mathrm{298 \mathrm{~K} :}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B298%20%5Cmathrm%

Calculate the standard Gibbs free energy change for the following reaction at \mathrm{298 \mathrm{~K} :}

\mathrm{ 2 \mathrm{H}_2(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l) }

Given the following data:

\mathrm{\begin{tabular}{|c|c|c|} \hline Compound & Standard enthalpy of formation $(\mathrm{kJ} / \mathrm{mol})$ & Standard entropy $(\mathrm{J} / \mathrm{K} \mathrm{mol})$ \\ \hline $\mathrm{H} 2$ & 0 & 130.7 \\ $\mathrm{O} 2$ & 0 & 205.0 \\ $\mathrm{H} 2 \mathrm{O}$ & -285.8 & 69.9 \\ \hline \end{tabular}}

Option: 1

\mathrm{-475 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{475 / \mathrm{mol}}


Option: 3

\mathrm{600 \mathrm{~kJ} / \mathrm{mol}}


Option: 4

\mathrm{-600 \mathrm{~kJ} / \mathrm{mol}}


Answers (1)

best_answer

The standard Gibbs free energy change of the reaction is given by:

                          \mathrm{ \Delta G_{r x n}^{\circ}=\sum \nu_i \Delta G_{f, i}^{\circ} }

where \mathrm{\nu_i} is the stoichiometric coefficient of species i and \mathrm{\Delta G_{f, i}^{\circ}} is the standard Gibbs free energy of formation of species i. Using the data given, we can write:

\mathrm{ \Delta G_{r x n}^{\circ}=\left(2 \times \Delta G_{f, H_2 O}^{\circ}\right)-\left(2 \times \Delta G_{f, H_2}^{\circ}\right)-\Delta G_{f, O_2}^{\circ} }

where \mathrm{\Delta G_{f, \mathrm{H}_2 \mathrm{O}}^{\circ}, \Delta G_{f, \mathrm{H}_2}^{\circ}}, and \mathrm{\Delta G_{f, \mathrm{O}_2}^{\circ}} are the standard Gibbs free energies of formation of water, hydrogen, and oxygen, respectively. We can calculate these values using the following equations:

                           \mathrm{ \begin{gathered} \Delta G_{f, \mathrm{H}_2 \mathrm{O}}^{\circ}=-T \Delta S_{f, \mathrm{H}_2 \mathrm{O}}^{\circ}+\Delta H_{f, \mathrm{H}_2 \mathrm{O}}^{\circ} \\\\ \Delta G_{f, \mathrm{H}_2}^{\circ}=-T \Delta S_{f, \mathrm{H}_2}^{\circ}+\Delta H_{f, \mathrm{H}_2}^{\circ} \\\\ \Delta G_{f, \mathrm{O}_2}^{\circ}=-T \Delta S_{f, \mathrm{O}_2}^{\circ}+\Delta H_{f, \mathrm{O}_2}^{\circ} \end{gathered} }
where T is the temperature in Kelvin,\Delta S_f^{\circ} is the standard entropy of formation, and \Delta H_f^{\circ} is the standard enthalpy of formation. Substituting the values given, we get:

                                \mathrm{ \begin{gathered} \Delta G_{r x n}^{\circ}=(2 \times(-237.13 \mathrm{~kJ} / \mathrm{mol}))-(2 \times 0)-0 \\\\ \Delta G_{r x n}^{\circ}=-474.26 \mathrm{~kJ} / \mathrm{mol} \end{gathered} }
Therefore, the standard Gibbs free energy change for the reaction is \mathrm{-474.26 \, \, \mathrm{kJ} / \mathrm{mol} \, \, at \, \, 298 \mathrm{~K}.}
So, the correct option is (A)

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 Registration LIVE: NTA session 1 form link at jeemain.nta.nic.in; application deadline
anam-khan

JEE 2026: IIT admissions show major shift toward AI, data science, away from core engineering courses
anam-khan

JEE Main 2026: NIT Delhi cut-offs for BTech courses show ever-increasing demands for AI, computer skills

Latest Question