Question

(Review Topics] (References Use the References to access important values if needed for this question. For the reaction HS() + 2H20(1) 3H2(g) + S029) AH° = 295 kJ and AS° = 295J/K AGº for this reaction would be negative at temperatures (above, below) Enter above or below in the first box and enter the temperature in the second box. Assume that A Hº and AS are constant. Submit Answer R etry Entire Group 2 more group attempts remaining

Answer 1

We use the equation: $\Delta$ G = $\Delta$ H - T$\Delta$S

We obtain the temperarture upto which / above which the reaction $\Delta$ G is negative byb equating the Right side of the equation to 0

1. $\Delta$ H = 295 KJ

$\Delta$S = 295 J/K

Using in the euqtaion: $\Delta$ H - T$\Delta$S = 0

295 * 10³ - T * 295 = 0

Solving, T = 1000 K

Therefore ABOVE, 1000 K G is negative

2.

$\Delta$H = 178 KJ

$\Delta$S = 161 J/K

Using in the euqtaion: $\Delta$ H - T$\Delta$S = 0

178 * 10³ - T * 161 = 0

SOlving, T = 1105.6 K

Therefore ABOVE, 1105.6 K G is negative