Question

7- (15 pts) Calculate the standard enthalpy of formation of CS2 (1) given that a. C(s) + O2(g) + CO2(g) b. S(s) + O2(g) → SO2(e) C. CS2() + 302(g) → CO2(g) + ' 250 (6) AH° = -393.5 kJ/mol AH° = -296.4 kJ/mol AH° = -1073.6 kj/mol (the standard state of the element Sulfur is S(s))

Answer 1

(45 vеh 90o С? өц - a. C(s) + O₂ g co a AH = -393.5 kJ/mol b. SS) & 02 502 (9) ON = -296.4 kg/rul 32 (1) 302 (g) > CO2 (g) + 250g (g) All = -107 3.6 k/oner The reaction for standard enthalpy of formation of ls 2 is governs as ! - C & 25cs a CS2 W we have to calculate the AH value for the above reachon using the values of the given reactions.

of the reach on (a) is added to 2x reaction (6) (d) C (s) & O2 (g) + 25 (s) + 201g) coz cg + 2502 lgs A = AH cas + oH (b) x2 = -393.5+ (296.4) x2 -393.5 - 592.8 -986.3 kJ fool Now " added the to reaction reaction (c) e (d), the reversed reaction then we get :- co₂ g + 2502 (g) + C(s) + O2(g) + 25 (s) + 2 029) - CS₂ W +32(g) t co2g + 2502 cg) ou simplifiration or Cast 25.6) - CS2 (d) Hi = A HOHOH (c) Reaction c' 9 2 -9865+ (107 3.6) was reversed its 2 + 87.3 KI/mol. Allvalve will also become positive from I negative)