Given:
Hof(CH4(g)) = -74.87 KJ/mol
Hof(Cl2(g)) = 0.0 KJ/mol
Hof(CCl4(l)) = -139.0 KJ/mol
Hof(HCl(g)) = -92.31 KJ/mol
Balanced chemical equation is:
CH4(g) + 4 Cl2(g) ---> CCl4(l) + 4 HCl(g)
ΔHo rxn = 1*Hof(CCl4(l)) + 4*Hof(HCl(g)) - 1*Hof( CH4(g)) - 4*Hof(Cl2(g))
ΔHo rxn = 1*(-139.0) + 4*(-92.31) - 1*(-74.87) - 4*(0.0)
ΔHo rxn = -433.37 KJ
Answer: -433.37 KJ
Calculate AH® for the following: rxn CH4(g) + Cl2(g) → CCI_(1) + HCl(g) (unbalanced] NH (CH4()]...
Calculate AH® for the following: rxn CH4(8) + Cl2(g) → CCl4(!) + HCl(g) [unbalanced] -139 kJ/mol AH (CH4®= -74.87 kJ/mol a CCl4@)) = -96.0 kJ/mol AHCC140] =-139 kJ/mol AH [HCI@)) =-92.31 kJ/mol AH [HCl(aq)] =-167.46 kJ/mol Ahº [C1()] = 121.0 kJ/mol
Enter your answer in the provided box. Calculate AH for the following: rxn CH )+Cl2)-CCL)+ HCIg) [unbalanced] AH [CH (g)]=-74.87 kJ/mol f AH [CCI(g)]-96.0 kJ/mol f AH [CCI () -139 kJ/mol f AH HClg)] = -92.31 kJ/mol f AH, [HCl(aq)]=-167.46 kJmol AH, [CIg))= 121.0 kJmol kJ
3 attempts left Check my work Enter your answer in the provided box Calculate AH nxn for the following: points eBook CH4(%) + Ch) CC14() + HCl() (unbalanced] Ask Print References AH ICH,®) =-74.87 kJ/mol AH CCu(e) = -96.0 kJ/mol AH (CC140) = -139 kJ/mol AH (HCI@)] = -92.31 kJ/mol AH (HCl(aq)] = -167.46 kJ/mol AH (C1(2)) = 121.0 kJ/mol Prev 8 of 15 H N
Part A Use the information provided to determine AH°rxn for the following reaction: AHⓇ CH4(9) + 4 Cl2(g) → CCl4(g) + 4 AH°rxn = (kJ/mol) HCl(9) CH4(9) -75 CCl4(g) -96 HCl(g) -92 +79 kJ O -389 kJ 0 -113 kJ O +113 kJ O -71 kJ Submit Request Answer
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