The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$. The $C-H$ bond enthalpy of each successive step in $CH_4$ $\rightarrow CH_3$ $\rightarrow CH_2$ $\rightarrow CH$ are $+15, +30$ and $+45 \ kJ \ mol^{-1}$ higher than the mean bond enthalpy of $C-H$ bonds,respectively. The bond enthalpy of the last $C-H$ unit is
- A$400 \ kJ \ mol^{-1}$
- B$325 \ kJ \ mol^{-1}$
- C$475 \ kJ \ mol^{-1}$
- D$385 \ kJ \ mol^{-1}$
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Calculate the heat of formation of $PCl_{5(s)}$ from the following data:
$2P_{(s)} + 3Cl_{2(g)} \to 2PCl_{3(l)}; \Delta H_1 = -151.8 \ kcal$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)}; \Delta H_2 = -32.8 \ kcal$
Determine the enthalpy of formation for $PCl_{5(s)}$ in $kcal$.
$2P_{(s)} + 3Cl_{2(g)} \to 2PCl_{3(l)}; \Delta H_1 = -151.8 \ kcal$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)}; \Delta H_2 = -32.8 \ kcal$
Determine the enthalpy of formation for $PCl_{5(s)}$ in $kcal$.
Medium
View SolutionThe enthalpy change for a reaction does not depend on:
Easy
View SolutionFind the value of $Q$ from the following equations:
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
The heat of atomisation of methane and ethane are $x \ kJ \ mol^{-1}$ and $y \ kJ \ mol^{-1}$ respectively. The longest wavelength $(\lambda)$ of light capable of breaking the $C-C$ bond can be expressed in $SI$ unit as :
DifficultJEE MAIN 2026
View SolutionGiven:
$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H^o_{298\ K} = -285.9 \ kJ \ mol^{-1}$
$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H^o_{298\ K} = -241.8 \ kJ \ mol^{-1}$
The molar enthalpy of vaporisation of water will be $kJ \ mol^{-1}$.
$(I) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H^o_{298\ K} = -285.9 \ kJ \ mol^{-1}$
$(II) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H^o_{298\ K} = -241.8 \ kJ \ mol^{-1}$
The molar enthalpy of vaporisation of water will be $kJ \ mol^{-1}$.
DifficultJEE MAIN 2013
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