Chemistry - CHEMICAL BONDING Question with Solution | TestHub

ChemistryCHEMICAL BONDINGMOTHard2 minPYQ_2012

ChemistryHardsingle choice

Bond enthalpies of $H_{2}, X_{2}$ and $HX$ are in the ratio 2: 1: 2 . If enthalpy of formation of $HX$ is $- 50 kJ$ $mol^{- 1},$ the bond enthalpy of $X_{2}$ is

Options:

Answer:

Solution:

$\frac{1}{2} H_{2} + \frac{1}{2} X_{2} ⟶ HX$ Let the bond enthalpy of $X - X$ bond be $x$ . $Δ H_{f} (HX) = - 50$ $= \frac{1}{2} Δ H_{H - H} + \frac{1}{2} Δ H_{X - X} - Δ H_{H - X}$ $= \frac{1}{2} 2 x + \frac{1}{2} x - 2 x = \frac{- x}{2}$ $∴ x = 50 \times 2 = 100 kJ mol^{- 1}$

Stream:BITSATSubject:ChemistryTopic:CHEMICAL BONDINGSubtopic:MOT

⏱ 2mℹ️ Source: PYQ_2012

Doubts & Discussion

Loading discussions...