Q.56. Consider the following data: (methane, g) = -X kJ mol Enthalpy of sublimation of graphite = Y

Q: What is the correct answer for JEE Main Chemistry Q56?

The correct answer is A. We need to find the average C-H bond enthalpy. Consider the formation of methane from its elements in their standard states: C(graphite) + 2H$_2$(g) $\rightarrow$ CH$_4$(g). The enthalpy of formation of methane is given as $\Delta H_f^{\circ}(\text{CH}_4) = -X$ kJ mol$^{-1}$. We can also calculate $\Delta H_f^{\circ}(\text{CH}_4)$ using bond enthalpies and sublimation enthalpy. The process can be

Q: Q.56. Consider the following data: (methane, g) = -X kJ mol Enthalpy of sublimation of graphite = Y kJ mol Dissociation enthalpy of H = Z kJ mol The bond enthalpy of C – H bond is given by:

The correct answer is A. We need to find the average C-H bond enthalpy. Consider the formation of methane from its elements in their standard states: C(graphite) + 2H$_2$(g) $\rightarrow$ CH$_4$(g). The enthalpy of formation of methane is given as $\Delta H_f^{\circ}(\text{CH}_4) = -X$ kJ mol$^{-1}$. We can also calculate $\Delta H_f^{\circ}(\text{CH}_4)$ using bond enthalpies and sublimation enthalpy. The process can be

Q: What are the options for this JEE Main Chemistry question?

A. $\frac{X+Y+2Z}{4}$ · B. $\frac{X+Y+4Z}{2}$ · C. X + Y + Z · D. $\frac{-X+Y+Z}{4}$

Question

Q.56. Consider the following data:  (methane, g) = -X kJ mol Enthalpy of sublimation of graphite = Y kJ mol Dissociation enthalpy of H = Z kJ mol The bond enthalpy of C – H bond is given by:

FirstInTest · Accepted Answer

We need to find the average C-H bond enthalpy. Consider the formation of methane from its elements in their standard states: C(graphite) + 2H$_2$(g) $ightarrow$ CH$_4$(g). The enthalpy of formation of methane is given as $\Delta H_f^{\circ}(	ext{CH}_4) = -X$ kJ mol$^{-1}$. We can also calculate $\Delta H_f^{\circ}(	ext{CH}_4)$ using bond enthalpies and sublimation enthalpy. The process can be broken down into steps: 1. Sublimation of graphite: C(graphite) $ightarrow$ C(g). Energy required = Y kJ mol$^{-1}$. 2. Dissociation of hydrogen molecules: 2H$_2$(g) $ightarrow$ 4H(g). Energy required = $2 	imes Z$ kJ mol$^{-1}$ (since Z is for one H$_2$ molecule, and we have two). 3. Formation of methane from gaseous atoms: C(g) + 4H(g) $ightarrow$ CH$_4$(g). Energy released = $4 	imes E_{	ext{C-H}}$, where $E_{	ext{C-H}}$ is the average C-H bond enthalpy. So, $\Delta H_f^{\circ}(	ext{CH}_4) = Y + 2Z - 4 E_{	ext{C-H}}$. We are given $\Delta H_f^{\circ}(	ext{CH}_4) = -X$. So, $-X = Y + 2Z - 4 E_{	ext{C-H}}$. Rearranging to find $E_{	ext{C-H}}$: $4 E_{	ext{C-H}} = Y + 2Z + X$. $E_{	ext{C-H}} = \frac{X + Y + 2Z}{4}$.

Answer

$\frac{X+Y+2Z}{4}$

Answer

$\frac{X+Y+4Z}{2}$

Answer

X + Y + Z

Answer

$\frac{-X+Y+Z}{4}$

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