The cell in which the following reaction occurs has at 298 K. The standard Gibbs energy will be

At 298 K, H0 = -314 kJ/mol and S0 = -0.372 kJ/(K•mol). What is the Gibbs free energy of the reaction?A.-425 kJB.0.393 kJC.34,900 kJD.-203 kJ

the standard electrode potential of a daniell cells is 1.1V. find the standard gibbs energy for the reaction

Using the following data at 298 K:Substance Df H° [kJ mol-1 ] S° [J mol-1 K -1 ]H 2 (g) 0 130.7CO (g) -110.5 197.7H 2 CO (g) -108.6 218.8(i) [2 marks]Calculate the standard Gibbs free energy change for the reaction at 298 K:H 2 (g) + CO (g) ® H 2 CO (g)

Calculate the Gibbs free energy for formation of C2H4. Comment on the spontaneityor otherwise of the reaction at 298K.Given 2C(s)+ 2H2(g)→C2H4(g); ∆H298K = 8Kcal.Substance So(in cal.deg-1mole-1)C(s) 1.4H2(g) 31.2C2H4(g) 52.2

The reaction below represents the decomposition of a generic diatomic molecule in its standard state.1/2 A2(g) → A(g)Assuming that the standard molar Gibbs energy of formation of A(g) is 8.12 kJ•mol–1 at 2500 Kand –72.20 kJ•mol–1 at 3500 K, determine the value of ΔH°rxn for these data assuming that ΔH°rxn is independent of temperature. Express your answer in kJ, in one decimal place.