Given that bond energy of H—H and Cl-Cl is 430 kJ mol^-1 and 240 kJ mol^-1 respectively and H_f for HCl is -90 kJ mol^-1 . The bond enthalpy of HCl is - Chemistry MCQ Chemistry | ExamTest.live

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Given that bond energy of H—H and Cl-Cl is 430 kJ mol^{- 1} and 240 kJ mol^{- 1} respectively and Δ H_{f} for HCl is - 90 kJ mol^{- 1} . The bond enthalpy of HCl is:

A $290$
B $380$
C $425$
D $245$

Solution & Step-by-step Explanation

The formation reaction is: \frac{1}{2} H_{2} (g) + \frac{1}{2} Cl_{2} (g) \to HCl (g) Δ H_{f} = \sum B.E. (reactants) - \sum B.E. (products) - 90 = [\frac{1}{2} B.E.(H-H) + \frac{1}{2} B.E.(Cl-Cl)] - B.E.(HCl) - 90 = [\frac{1}{2} (430) + \frac{1}{2} (240)] - B.E.(HCl) - 90 = [215 + 120] - B.E.(HCl) - 90 = 335 - B.E.(HCl) B.E.(HCl) = 335 + 90 = 425 kJ mol^{- 1} .

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Given that bond energy of H—H and Cl-Cl is 430 kJ mol^{- 1} and 240 kJ mol^{- 1} respectively and Δ H_{f} for HCl is - 90 kJ mol^{- 1} . The bond enthalpy of HCl is:

290

380

425

245

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Thermodynamics Bond Enthalpy

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Given that bond energy of $H—H$ and $Cl-Cl$ is $430 kJ mol^{- 1}$ and $240 kJ mol^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- 90 kJ mol^{- 1}$ . The bond enthalpy of $HCl$ is:

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Discussion

Given that bond energy of H—H and Cl-Cl is 430 kJ mol−1 and 240 kJ mol−1 respectively and ΔHf​ for HCl is −90 kJ mol−1. The bond enthalpy of HCl is:

Solution & Step-by-step Explanation

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Discussion

Given that bond energy of $H—H$ and $Cl-Cl$ is $430 kJ mol^{- 1}$ and $240 kJ mol^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- 90 kJ mol^{- 1}$ . The bond enthalpy of $HCl$ is: