neren
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Seorang penemu pikir ia telah mengembangkan katalis baru yang dapat reaksi fase gas CO 2 + 4H 2 2H 2 O + CH 4 Memproses 100% konversi, T 1 500 ◦ C, hitung T 2 jawab : tabulated data CO 2 4H 2 2H 2 O CH 4 - ∆Hf ◦ (J/g mol) 393.513 0 241.826 74.848 ∆H rxn298K = [-74.848 – 2(241.826)] – [4(0) – 393.513] = -164.987 J/g mol CO 2 ∆H p - ∆H rx Cp CO 2 = 26,75 + 42,26(10 -3 )T - 14,25(10 -6 )T 2 Cp H 2 = 26,88 + 4,35(10 -3 )T – 0,33(10 -6 )T 2 Cp H 2 O = 29,16 + 14,49(10 -3 )T – 2,02(10 -6 )T 2 Cp CH 4 = 13,41 + 77,03(10 -3 )T – 18,74(10 -6 )T 2 ∆a = [1(13,41) + 2(29,16) – 1(26,75) + 4(26,88)] = -62,54 ∆b = [1(77,03(10 -3 )) + 2(14,49(10 -3 )) – 1(42,26(10 -3 )) + 4(4,35(10 -3 ))] = 46,35(10 -3 ) ∆c = [1(-18,74(10 -6 )) + 2(-2,02(10 -6 )) – 1(-14,25(10 -6 )) + 4(- 0,33(10 -6 ))] = -7,21(10 -6 ) ∆H p - ∆H rx = 773 ∫ T [-62,54 + 46,35(10 -3 )T -7,21(10 -6 ) T 2 ] dT ∆H p - ∆H rx = -62,54T + 46,35(10 -3 )/2 T 2 -7,21(10 -6 )/3 T 3 ∆H p - ∆H rx = -62,54T + 23,175 (10 -3 ) T 2 – 2,4003(10 -6 ) T 3 ∆H p - ∆H rx = -62,54T + 23,175 (10 -3 ) T 2 – 2,4003(10 -6 ) T 3 ∆H p - ∆H rx = -62,54T + 23,175 (10 -3 ) T 2 – 2,4003(10 -6 ) T 3
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Seorang penemu pikir ia telah mengembangkan katalis baru yang dapat reaksi fase gasCO2 + 4H2 2H2O + CH4Memproses 100% konversi, T1 500C, hitung T2jawab :tabulated dataCO2 4H2 2H2O CH4- Hf (J/g mol)393.5130241.82674.848Hrxn298K = [-74.848 2(241.826)] [4(0) 393.513]= -164.987 J/g mol CO2Hp - HrxCp CO2 = 26,75 + 42,26(10-3)T - 14,25(10-6)T2 Cp H2 = 26,88 + 4,35(10-3)T 0,33(10-6)T2Cp H2O = 29,16 + 14,49(10-3)T 2,02(10-6)T2Cp CH4 = 13,41 + 77,03(10-3)T 18,74(10-6)T2a = [1(13,41) + 2(29,16) 1(26,75) + 4(26,88)]= -62,54b= [1(77,03(10-3)) + 2(14,49(10-3)) 1(42,26(10-3)) + 4(4,35(10-3))]= 46,35(10-3)c= [1(-18,74(10-6)) + 2(-2,02(10-6)) 1(-14,25(10-6)) + 4(-0,33(10-6))]= -7,21(10-6)Hp - Hrx= 773T [-62,54 + 46,35(10-3)T -7,21(10-6) T2] dTHp - Hrx = -62,54T + 46,35(10-3)/2 T2 -7,21(10-6)/3 T3Hp - Hrx = -62,54T + 23,175 (10-3) T2 2,4003(10-6) T3Hp - Hrx = -62,54T + 23,175 (10-3) T2 2,4003(10-6) T3Hp - Hrx = -62,54T + 23,175 (10-3) T2 2,4003(10-6) T3
Hrxn298K = Hp - Hrx-164.987 = -62,54T + 23,175 (10-3) T2 2,4003(10-6) T30= -62,54T + 23,175 (10-3) T2 2,4003(10-6) T3 + 164.987T= T2= T 773T2=
