Calculate the standard enthalpy change (in kJ mol-1 ) for the reaction H2(g)+O2(g) xrightarrowH2O2(g) Given that bond enthalpies of H-H, text O=O, OH and OO (in kJ mol-1 ) are respectively 438, 498, 464 and 138.

Question

Calculate the standard enthalpy change (in kJ mol-1 ) for the reaction H2(g)+O2(g) xrightarrowH2O2(g) Given that bond enthalpies of H-H, text O=O, OH and OO (in kJ mol-1 ) are respectively 438, 498, 464 and 138. (A)  -130  (B)  - 65  (C) 130 (D)  - 334

Tardigrade Admin · Accepted Answer

H2(g)+O2(g) xrightarrowH2O2(g) Δ Hreaction=BEreac tan ts-BEproducts =[BE(H-H)+BE(O=O)] -2BE(O-H)+BE(O-O)] =[438+498]-[2× 464+138] =936-1066=-130 text kJ text mol-1

Q. Calculate the standard enthalpy change (in $k J m o l^{- 1}$ ) for the reaction
$H_{2} (g) + O_{2} (g) H_{2} O_{2} (g)$
Given that bond enthalpies of $H - H, O = O,$ $O H$ and $OO$ (in kJ $m o l^{- 1}$ ) are respectively $438, 498, 464$ and $138$ .

$- 130$

$- 65$

130

$- 334$

334

$H_{2} (g) + O_{2} (g) H_{2} O_{2} (g)$
$Δ H_{re a c t i o n} = B E_{re a c t a n t s} - B E_{p ro d u c t s}$
$= [BE (H - H) + BE (O = O)]$
$- 2 BE (O - H) + BE (O - O)]$
$= [438 + 498] - [2 \times 464 + 138]$
$= 936 - 1066 = - 130 k J m o l^{- 1}$

Q. Calculate the standard enthalpy change (in kJmol−1 ) for the reaction H2​(g)+O2​(g)​H2​O2​(g) Given that bond enthalpies of H−H, O=O, OH and OO (in kJ mol−1 ) are respectively 438,498,464 and 138.

−130

−65

130

−334

334

H2​(g)+O2​(g)​H2​O2​(g) ΔHreaction​=BEreactants​−BEproducts​ =[BE(H−H)+BE(O=O)] −2BE(O−H)+BE(O−O)] =[438+498]−[2×464+138] =936−1066=−130kJmol−1

Q. Calculate the standard enthalpy change (in $k J m o l^{- 1}$ ) for the reaction
$H_{2} (g) + O_{2} (g) H_{2} O_{2} (g)$
Given that bond enthalpies of $H - H, O = O,$ $O H$ and $OO$ (in kJ $m o l^{- 1}$ ) are respectively $438, 498, 464$ and $138$ .

$- 130$

$- 65$

$- 334$

$H_{2} (g) + O_{2} (g) H_{2} O_{2} (g)$
$Δ H_{re a c t i o n} = B E_{re a c t a n t s} - B E_{p ro d u c t s}$
$= [BE (H - H) + BE (O = O)]$
$- 2 BE (O - H) + BE (O - O)]$
$= [438 + 498] - [2 \times 464 + 138]$
$= 936 - 1066 = - 130 k J m o l^{- 1}$