Question

Then, he measures the reaction enthalpy AH and reaction entropy AS of the second reaction. The results of his measurements are shown in the table. Complete the table. That is, calculate AG for the first reaction and AS for the under the conditions the engineer has set up, the reaction is spontaneous, the spontaneous because the system is at equilibrium. AH = -188. kJ AS = -602. AG - kJ N,H,() + H2(g) — 2NH3(e) Which is spontaneous? this reaction the reverse reaction neither AH = -1353. kJ as - 0 2CH2OH(g) + 30, (g) → 2002(g) + 4H,0 (8) AG = -11. kJ Which is spontaneous? this reaction the reverse reaction neither

Answer 1

For first reaction : We know

$\Delta G =\Delta H-T\Delta S$

$\Delta H$ = -188 kJ

$\Delta S$ = -602 J/K = -0.602 kJ/K

T = 39oC = 312 K

$\Delta G$ = (-188) - (312)(-0.602)

$\Delta G$ = -0.176 kJ

As $\Delta G$ is negative .So The given reaction is spontaneous .Option A is correct.

For second reaction :

$\Delta G$ = -11 kJ

$\Delta H$ = -1353 kJ

T =312 K

$\Delta S$ = -[(-11+1353)/312]

= -4.301 kJ / K = -4301 J/K

As $\Delta G$ is negative .So The given reaction is spontaneous .Option A is correct.