InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The standard heat of formation values of `SF_(6)(g), S(g)`, and `F(g)` are `-1100, 275`, and `80 kJ mol^(-1)`, respectively. Then the average `S-F` bond enegry in `SF_(6)`A. `310 kJ mol^(-1)`B. `220 kJ mol^(-1)`C. `309 kJ mol^(-1)`D. `280 kJ mol^(-1)` |
|
Answer» `S(g) +6F(g) rarr SF_(6)(g), DeltaH =- 1100 kJ mol^(-1)` `S(s) rarr S(g) ,DeltaH =+ 275 kJ mol^(-1)` `(1)/(2) F_(2)(g) rarr F(g),DeltaH = 80 kJ mol^(-1)` Therefore, heta of formation =Bond enegry of reaction -Bond enegry of product `- 1100 = (275 +6 xx 80) - (6 xx S -F)` Thus, bond enegry of `S-F = 309 kJ mol^(-1)` |
|