Given, `S_((s))+3F_(2(g))rarr SF_(6),DeltaH=-1100kJ ….(1)`
`S_((s)) rarr S_((g)),DeltaH=275kJ …(2)`
`(1)/(2)F_(2(g))rarr F_((g)), DeltaH=80kJ ..(3)`
To get `SF_(6(g))rarr S_((g))++6F_((g)),,`
we can proceed as `eq. (2)+6xxeq.(3)-eq.(1)`
`SF_(6(g))rarrS_((g))+6F_((g)),DeltaH=1855kJ`
Thus, average bond energy `=(1855)/(6)=309.17`
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)`
Abhinav Prasad Wali
Asked: 3 years ago2022-11-01T20:20:32+05:30
2022-11-01T20:20:32+05:30In: General Awareness
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)`
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)`
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Fakaruddin Shenoy
Asked: 3 years ago2022-10-29T01:23:57+05:30
2022-10-29T01:23:57+05:30In: General Awareness
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)`
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)`
A. `310 kJ mol^(-1)`
B. `220 kJ mol^(-1)`
C. `309 kJ mol^(-1)`
D. `280 kJ mol^(-1)`
Leave an answer
Leave an answer
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`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)`
Correct Answer – 309.16 kJ