For which ONE of the following substances is the standard free energy of formati

Question

For which ONE of the following substances is the standard free energy of formation (DeltaG_f*) not equal to zero at 298 K? A) Ne(g) B) N_2(g) C) O_2(g) D) O_3(g) E) Cu(s) 10. A certain reaction has negative values for both DeltaH and DeltaS. Therefore, which ONE of the statements below MUST BE correct? A) The reaction can be spontaneous if the temperature is low enough. B) The reaction cannot be spontaneous at any temperature. C) The reaction must be spontaneous at all temperatures. D) The reaction can be spontaneous if the temperature is high enough. E) The reaction has a positive free energy at any temperature. 11. The standard free energy change for a chemical reaction is -6.73 kJ/mol. What is the equilibrium constant for the reaction at 25degree C? A) 0.0019 B) 0.066 C) 1.0 D) 15 E) 520 12. Which ONE of the following reactions is NOT an oxidation-reduction reaction? A) H_2(g) + F_2(g) rightarrow 2 HF(g) B) Ca(s) + H_2(g) rightarrow CaH_2(s) C) 2K(s) + 2H_2O(l) rightarrow 2KOH(aq) + H_2(g) D) 6 Li(s) + N_2(g) rightarrow 2 Li_3N(s) E) Mg_3N_2(s) + 6 H_2O(l) rightarrow 3 Mg(OH)_2(s) + 2 NH_3(g) 13. The oxidation number of nitrogen given for the following compounds is correct except for which ONE? A) hydrazine, N_2H_4 (-2) B) sodium acide, NaN_3 (-1) C) nitrous acid, HNO_2 (+3) D) dinitrogen trioxide, N_2O_3 (+3) E) nitramide, H_2N_2O_2 (+1) 14. Which equation is the balanced half-reaction for the reduction of hydrogen peroxide (H_2O_2) to water in an acidic solution? I A) 2H_2O_2(l) rightarrow 2 H_2O(l) + O_2(g) B) 2 H_2O_2(l) + 2e^- rightarrow 2 H_2O(l) + O_2(g) C) H_2O_2(l) + 2 H^+(aq) + 2 e^- rightarrow 2 H_2O(l) D) H_2O_2(l) + 4 H^+(aq) + 2 e^- rightarrow 2 H_2O(l) + H_2(g) E) H_2O_2(l) + 2 H^+(aq) + 4e^- rightarrow 2 H_2(g) + O_2(g)

Explanation / Answer

9.

Fo O3, since it is not in its elemental state (i.e. that will be O2(g))

10.

if

G = H- TS

and both are negative, then this is inly favoured at LOW T; if T is high, then G > 0, which will make it possitve

11.

G = -RT*lnK

K = xp(-G/(RT) = exp(6730/(8.314*298) = 15.12530

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