Question

Need help with #8 & 10. If they are correct will rate high. Thank you!

8. The Haber process for the production of ammonia involves the equilibrium 2 NH3(g) → N2(g) + 3 H2(g) For this reaction, AH° = 92.38 kJ and AS = 198.3 J/K. Assume that AH° and AS° for this reaction do not change with temperature. (i) Predict the direction in which AG for the reaction changes with increasing temperature. (ii) Calculate AG at 25 °C and at 500 10. The standard free-energy change for the Haber process at 25°C was obtained in Sample Exercise 19.9 for the Haber reaction: N2(g) + 3 H2(g) → 2 NH3(a)_ AG° =-31.3 kJ/mol Use this value of AGº to calculate the equilibrium constant for the process at 25 °C. a. 6.87 x 105 b. 7.89 x 102 c. 9.05 x 108 d. 2.57 x 100 e. 3.07 x 105 •C a. Less negative; 33.3 kJ, 61 kJ b. more negative; -33.3 kJ, 61 kJ C. Less negative;-61 k), -61 kJ d. more negative; 33.3 kJ, -61 kJ e. Less negative; 161 kJ, 33.3 kJ

Answer 1

③ - The given equilibrium process is & NHGg. 7 Nang + 3 hengo The relation between AG, Dho and A8 given by AG DH_TAS → is (the) Given. DH₂92; 38 kJ ? Both are positive As - 198.3 *5|k] V DH and so Both are positive, if the "temperature (I)" increases the the value of as increases more, thens "AG-value is more negetive (ve), because 'I' multiplies with "AS" ie, from equation- AG AH (TAS) This value increases -

of it, then G is - with more increase negetine 6 مطالبه ای ها و مع له Given, DHO. 92.38 KJ , DS 198.3 TIK (:15-1õØKJ] s8=198.3 x 103 kJ/k T = 25C - (25+873)k z 898k AG = DH- TAS AG: 98.38 KJ - 298k * 198.3x10x1 Then. OG: 98.38kJ-[ 298 0.1983 kJ] AG 92.38 kJ – 59.09 kJ DG 2 33.89 kJ (or) 33.8kJ i AG at 850 33.3 kJ)

Calculation Given WA g DG at boot DH = 92.38 kJ (at 85% (or) 898k] evit per prom : - DG - 33.3 kJ Cat & SC ®) 8986] . (from above) Giat sood ? T, = 25C = (25+273)k - 898k Tę z 5000 = (500 + 273 )k 2 773x we know that, d(AG) = -(AHAT above equation between "T, ET la poo) :-( aholat . Integrate DE 97**-+

nowo, (+) Code - Carolina (te ] substitute values, (AG. Isaic Cavese. So take leone (oa luzie - 33.385 92:38 15 1736 2 ask! 773k &98k 773K 898k 713% (-0.190) KJİK - 0.0783 kg|K (OG)sac 773k 10G) gooc 7736 * -0.0783 kJ/k (OG)gooc - 60.58 kJ on -61 kJ/

.. Answer is, loption 18P3/ @ es more negetive, 33.3 kJ -61 kJ speto

10 Given equilibrium process is, Nag + 3 Hang ? &NH3 g, & free-energy Change (AG) = - 31-3 kJ land * Temperature (J). 256. (85 +273) K z 898K * gas - Constant (R) = 8.314 Jk mor The relation between AG & equilibrium Constant (K) is given by, G = - RT en k AG -RT (8:303 logk) dog klasenger Then, look -|- 31.3 kJ lma) &.303* 8:314J.K moi x2986

log k 3103 kJ 5705.8 J (1 kJ = 1035] logk 31.3x1039 5705. 85 logk 25.486 (loga za k a 105.486 10 k 23.06x105 Equilibrium Constant (K) = 3.06 x 100) - Answer is, option - ©: 3.07x105