(B) The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$.$CH_{4(g)} \longrightarrow C_{(g)} + 4H_{(g)} ; \Delta_a H = 1660 \ kJ \ mol^{-1}$Mea

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(B) The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$.$CH_{4(g)} \longrightarrow C_{(g)} + 4H_{(g)} ; \Delta_a H = 1660 \ kJ \ mol^{-1}$Mean $C-H$ bond enthalpy $= \frac{1660}{4} = 415 \ kJ \ mol^{-1}$.Let the bond enthalpies of the four $C-H$ bonds be $E_1, E_2, E_3, E_4$.Given: $E_1 = 415 + 15 = 430 \ kJ \ mol^{-1}$,$E_2 = 415 + 30 = 445 \ kJ \ mol^{-1}$,$E_3 = 415 + 45 = 460 \ kJ \ mol^{-1}$.The sum of all bond enthalpies equals the atomisation enthalpy:$E_1 + E_2 + E_3 + E_4 = 1660$$430 + 445 + 460 + E_4 = 1660$$1335 + E_4 = 1660$$E_4 = 1660 - 1335 = 325 \ kJ \ mol^{-1}$.

Class 11 Chemistry Thermodynamics Heat of reaction, Bond energy and Hess law

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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

AP EAMCET 2022 All AP EAMCET

A
$400 \ kJ \ mol^{-1}$
B
$325 \ kJ \ mol^{-1}$
C
$475 \ kJ \ mol^{-1}$
D
$385 \ kJ \ mol^{-1}$

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Class 11

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Thermodynamics

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Heat of reaction, Bond energy and Hess law

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Given that the bond energies of $N \equiv N$ is $946 \ kJ \ mol^{-1}$,$H-H$ is $435 \ kJ \ mol^{-1}$,$N-N$ is $159 \ kJ \ mol^{-1}$,and $N-H$ is $389 \ kJ \ mol^{-1}$,calculate the enthalpy of formation for the gas phase reaction $N_2 + 2H_2 \rightarrow N_2H_4$ in $kJ \ mol^{-1}$.

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The standard enthalpy of formation of $H_2O_{(g)}$ at $298 \ K$ is $-241.82 \ kJ \ mol^{-1}$. Calculate the enthalpy of formation of $H_2O_{(g)}$ at $373 \ K$,assuming $C_p$ is independent of temperature.
$C_p$ of $H_2O_{(g)} = 33.58 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $H_{2(g)} = 28.84 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $O_{2(g)} = 29.37 \ J \ K^{-1} \ mol^{-1}$

Medium

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Given: $C + O_2 \rightarrow CO_2$ : $\Delta H = -395 \ kJ$,$S + O_2 \rightarrow SO_2$ : $\Delta H = -295 \ kJ$,$CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$ : $\Delta H = -1110 \ kJ$. Calculate the heat of formation of $CS_2$ in $kJ/mol$.

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On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$

Medium

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The enthalpy change for the reaction,$C_{2}H_{6(g)} \to 2C_{(g)} + 6H_{(g)}$ is $X \ kJ$. The bond energy of $C-H$ bond is :-

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Given that the bond energies of $N \equiv N$ is $946 \ kJ \ mol^{-1}$,$H-H$ is $435 \ kJ \ mol^{-1}$,$N-N$ is $159 \ kJ \ mol^{-1}$,and $N-H$ is $389 \ kJ \ mol^{-1}$,calculate the enthalpy of formation for the gas phase reaction $N_2 + 2H_2 \rightarrow N_2H_4$ in $kJ \ mol^{-1}$.

Given: $C + O_2 \rightarrow CO_2$ : $\Delta H = -395 \ kJ$,$S + O_2 \rightarrow SO_2$ : $\Delta H = -295 \ kJ$,$CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$ : $\Delta H = -1110 \ kJ$. Calculate the heat of formation of $CS_2$ in $kJ/mol$.

On the basis of the following reactions,which one is correct?$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$

The enthalpy change for the reaction,$C_{2}H_{6(g)} \to 2C_{(g)} + 6H_{(g)}$ is $X \ kJ$. The bond energy of $C-H$ bond is :-

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On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$