given `NO(g)+O_(3)(g)toNO_(2)(g)+O_(2)(g) " "DeltaH=198.9kJ//mol` `O_(3)(g)to3//2O_(2)(g) " " DeltaH=-142.3 kj//mol` `O_(2)(g)to2O(g) " " DeltaH=+495.

Question

given `NO(g)+O_(3)(g)toNO_(2)(g)+O_(2)(g) " "DeltaH=198.9kJ//mol` `O_(3)(g)to3//2O_(2)(g) " " DeltaH=-142.3 kj//mol` `O_(2)(g)to2O(g) " " DeltaH=+495.

given `NO(g)+O_(3)(g)toNO_(2)(g)+O_(2)(g) " "DeltaH=198.9kJ//mol`
`O_(3)(g)to3//2O_(2)(g) " " DeltaH=-142.3 kj//mol`
`O_(2)(g)to2O(g) " " DeltaH=+495.0 kj//mol`
The entalpy change `(DeltaH)` for the following reaction is
`NO(g) + O (g) to NO_(2) (g)`
A. `-304.1 kJ//mol`
B. ` +304.1 kJ//mol`
C. `-403.1 kJ//mol`
D. `+403.1kJ//mol`

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - a
`NO+O_(3)toNO_(2)+O_(2) DeltaH=-198.9 "kJ//mole"`
`O_(3)to3/2O_(2)(g)DeltaH=-142.3 "Kj//mole"`
`O_(2)to2O_(2)(g)DeltaH=+495.0 "K J//mole"`
`DeltaH "of" NO+O(g)toNO_(2)(g)`
` "for this"i-ii-(iii)/2`
`198.9-(-142.3)-(495)/2`
304.1 " KJ//mole"`

6 views Answered 2023-02-05 15:29