Select the gas-phase reaction(s) which can pr | vedclass

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(D) reaction proceeds spontaneously if the ionization energy of the product species is lower than that of the reactant species,or if the electron affi

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$Kr + He^{+} 	o Kr^{+} + He$

Vedclass · Answer

(D) reaction proceeds spontaneously if the ionization energy of the product species is lower than that of the reactant species,or if the electron affinity of the product is higher than that of the reactant.In option $(D)$,$Kr + He^{+} 	o Kr^{+} + He$,the ionization energy of $Kr$ $(13.99 \ eV)$ is significantly lower than the ionization energy of $He$ $(24.58 \ eV)$.Since $He^{+}$ has a higher electron affinity (or ionization potential) than $Kr$,the electron will transfer from $Kr$ to $He^{+}$ to form the more stable $Kr^{+}$ and $He$ atoms.Thus,the reaction $Kr + He^{+} 	o Kr^{+} + He$ is spontaneous.

$I.E.$	$1^{st}$	$2^{nd}$	$3^{rd}$	$4^{th}$	$5^{th}$
Value	$577.5$	$1810$	$2750$	$11580$	$14820$

Select the gas-phase reaction$(s)$ which can proceed spontaneously?

Similar Questions

The first ionization enthalpies of $Na, Mg$ and $Si$ are respectively $496, 737$ and $786 \ kJ \ mol^{-1}$. What would be the first ionization enthalpy of $Al$ in $kJ \ mol^{-1}$?

The atomic number of the element from the following with the lowest $1^{st}$ ionisation enthalpy is:

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$I.E.$ $1^{st}$ $2^{nd}$ $3^{rd}$ $4^{th}$ $5^{th}$

Value $577.5$ $1810$ $2750$ $11580$ $14820$

The element is:

Which of the following orders for ionization energy is correct?

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