Question

For the reaction S(s,rhombic) + 2CO(g) SO2(g) + 2C(s,graphite) AH = -75.8 kJ and AS =-167.6J/K The standard free energy change for the reaction of 1.88 moles of S(s,rhombic) at 294 K, 1 atm would be This reaction is (reactant, product) favored under standard conditions at 294 K. Assume that AH° and AS are independent of temperature. Submit Answer Try Another Version 2 Item attempts remaining

Answer 1

Here the given the balanced reaction 5 (s, phombie) + 2 CO(g) — 50,(0) +20 (s, goaphat) Given, 4H = - 75.8 kJ = -75.8 x 1000 J. =-75800 L and, as = -167.6 J/k Given, temperature, T=294k Standard free energy can be obtained by the equation - AG = 4H - T. 45° a AG = -75800 et - 294 * x( 167.6) J. Kt =-75800 M + 4927 4.4 J =-26525.6 J - 26525.6 - KJ 1000 7 - 26.5 KT throee significant - figures ]

Now, for 1 mol of 5 (s, khombic) we get the standard free energy = -26.5 kJ 1 Thus for 1.88 moles of 5 (s, phombic) we I will get the standard free energy value = 1.88 x -26.5 kJ = - 49.8 kJ (throee significant I Thus, standard free energy change is 1 = -49.8 KJ Aus roee NO ИХК

As, 4G =-ve, the reaction is prooduct favoured under standard condition at 2944 as then forward process will be spon taneous.

I have given the answer in three significant figures.....if the answer does not match due to significant figures....then please comment below.....