Question

510). i. calculate the standard entalpy change for the reactichi Mg + % 0 Mgo Given the following dator Mg + 2HCl + Mg(12th, fH, L AH: -157:01 kJ AH MgO + 2H01 - MgCl₂ + H₂O | 1 Hº, -120 kg. H₂ + 1/ O H₂O A t= -286 kJ m. Calculate the enthalpy change for the reaction in C4 Hugo + Hng - Ca Horgh on coll Given the following date •CH (9) +302492.00249 +2H, Q. (1) • 2 C ₂ H 6 (g) + 7 0₂ (g) 7 4002 g + 6H₂O11 2 Harg) + Oarg 2H₂O cil A:1410.9 kJ AH = - 3 119.4 kJ A H=-571.6 J

Hi.can you help me answer 5)a)b)c).thank you so much for your help and kindness.I appreciate it

Answer 1

5(a)i) Given P + 2 H C = 10, 4 , 5 »'- -957.04 ky AO Mgo & The moc +40, SH - - 136 kJ y + 10 + 40 04 = -886 kJ To make me t o mgo Taking est equation as it is, reverse the equation 2, and third equation as it is, Overall reaction will be Mg + 40 mgel, + DH = -257.04 kJ O Mally tho MgO + 2Hill, DH = +186 kJ W + 10g - 40 , DH = - 286x5 ② Mg + 10 mo Overall enthalpy will be? DH = -257.dyk J + 126KJ - 286 kJ IDH = -417 01 K5 1) liven: Hy (9) + 302 (9) 2634 +70,(3) — 24, 19) + 0,19) 2(0,13) 434,0 , duº =-141.9 ks 410,4 610 , 04° = -3119.465 @ 24/1) DU = -171.6KT for the formation of Hylg) + (9) 0419) to reverse the 2nd and then dividend, and civice & byg. Au

Overall reaction will be Hy 9) + 30,19) —>346, (9)+ 38,0 V) , D#= -140.9 KT 21001+34011) - Cotilg) +70 g su= + 1559.7 kJ Hg (t) + 10919) MO(1) D4 = -285.8k5 Cg My(9) + Hy(s) — 648 (9) Overall enthalpy wh bez s4= 14 10. 9) KI + t 1559.7) Kg + (-285.8) kJ SP= -137 kJ Given doto :- (1418) + 20 219 2H₂ 011) + 10(9), DH = -891-45lmodo ((g) + O2(3) 10(9) , DH=-394 KJ /mel ③ 419) + f g - 4011) , DH = -286 kJ/mol for the formation of (19) +212 19) — 1419), reverse the , put egne as it is and multiply ear & by 2, and add all. Overall ean will be a 0,19) + 24,041) - CHy!9)+10,19), DH=841.ky (g) +0,1) 9819) , DH = - 394 kg 24,191 + 02/5) 2011), DU=.592 kr (19) + 21,69) CHy(9)

Overall Enthalpy will be or DH = (891) kJ/mol - (394) KJ/mol - (572) KT I mol [DH = -75 kJ lonel I 5(b) ) Criven data & 2 C₂ H₂ (g) + 5 Og 19) 4 (0 +24₂ 047 SH = -2599 kJ 0 CoMo (9) +70, 19) — 460904 6 ,00), 4° = -3119 kJ © 213 (9) + 0(g) - 2011) , 4 - - 572 KJ ③ for the formation of C. Ha (9) + 2 H219) - GHe 19 divide by a reverse the & and then divide by 2, & as it is and add all equations, Overall reaction will be: Ca Hg 19) + 5 0,19) 2008 (9) + 8), Du= - 1999.5 (0619 +3464) > M (9) + 3%13) , D10 = 41559.5 2 Mg (9) + 098) 21,018) Duo 578 kJ Coty (3) + 2 M319) (4219) Overall enthalpy will be - Dhe - (1299.5)*1 + (1554.5 kJ +(-578kr TSH = -312 kJ

S.C.1) for the formation of c (graphite ) + 2 H(g) + 302 /9) CH0H14) equation as it is, multiply & byd and neverse the 3 and add all. (t Og Cog , suº = -393.5 KT Imol 2H2 4.02 2H₂O SHO = 2x1-285.8) kJ/mol 20 + (0g – CH₂OH + 30 DH = 136.4 kJ/mol Overall 4* 2- C (graphite ) + 2 H+ og 14, OH overall enthalpy - DHO - (39315)KT Imol +(-571.69 KT/mal + 796.4*J) mal DHO -- 238.7 kJ lonel ] 5. Call for the formation of Mulg) + NH319) H(N(g) + 34(9), neverge en by dividing 2 , ear & is reversed and divide ean by 2, and add all, NH3 / Na + 3 H₂ , 04° = 445.9 kJ y & C + 24 DH = +74.9 KT Hg & C + + Ng HEN (Hylg) + NH₃(g) HON(g) + 3 H₂19) This is overall reaction. halpy Overall enthalpy & - DH = (45.9) kJ +14 Du= (45.9 kJ +(14.4) KT + 135.15 kJ TDH = 255.95 IDHO = + 135.15kJ