The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$? Given that the bond energy of $H_2$,$Br_2$,and $HBr$ is $435 \ kJ \ mol^{-1}$,$192 \ kJ \ mol^{-1}$,and $368 \ kJ \ mol^{-1}$ respectively.
- A$-109 \ kJ \ mol^{-1}$
- B$109 \ kJ \ mol^{-1}$
- C$259 \ kJ \ mol^{-1}$
- D$-259 \ kJ \ mol^{-1}$
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If $S + O_2 \to SO_2; (\Delta H = -298.2 \ kJ)$,$SO_2 + \frac{1}{2} O_2 \to SO_3; (\Delta H = -98.2 \ kJ)$,$SO_3 + H_2O \to H_2SO_4; (\Delta H = -130.2 \ kJ)$,$H_2 + \frac{1}{2} O_2 \to H_2O; (\Delta H = -287.3 \ kJ)$,then the enthalpy of formation of $H_2SO_4$ at $298 \ K$ will be......$kJ$.
Medium
View SolutionGiven the following thermochemical equations:
$C_{(s)} + O_{2_{(g)}} \to CO_{2_{(g)}} + 94.2 \, kcal$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \to H_2O_{(l)} + 68.3 \, kcal$
$CH_{4_{(g)}} + 2O_{2_{(g)}} \to CO_{2_{(g)}} + 2H_2O_{(l)} + 210.8 \, kcal$
Calculate the heat of formation of methane in $kcal$.
$C_{(s)} + O_{2_{(g)}} \to CO_{2_{(g)}} + 94.2 \, kcal$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \to H_2O_{(l)} + 68.3 \, kcal$
$CH_{4_{(g)}} + 2O_{2_{(g)}} \to CO_{2_{(g)}} + 2H_2O_{(l)} + 210.8 \, kcal$
Calculate the heat of formation of methane in $kcal$.
Medium
View SolutionObserve the following reaction:
$2 A_{2(g)} + B_{2(g)} \xrightarrow{T(K)} 2 A_2 B_{(g)} + 600 \ kJ$
The standard enthalpy of formation $(\Delta_f H^{\circ})$ of $A_2 B_{(g)}$ is:
$2 A_{2(g)} + B_{2(g)} \xrightarrow{T(K)} 2 A_2 B_{(g)} + 600 \ kJ$
The standard enthalpy of formation $(\Delta_f H^{\circ})$ of $A_2 B_{(g)}$ is:
The heat of combustion of $C_xH_y$,carbon,and hydrogen are $a, b$,and $c \ cal/mole$ respectively. The heat of formation of $C_xH_y$ will be:
Difficult
View SolutionWhen it is not possible to calculate the enthalpy of a reaction experimentally,it can be calculated by .....
Easy
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