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

Vedclass · Accepted Answer

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

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.

$(IP)_1$	$7.1 \ eV$
$(IP)_2$	$10.3 \ eV$
$(IP)_3$	$36.2 \ eV$
$(IP)_4$	$39.1 \ eV$

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

Similar Questions

Which statement is correct regarding ionisation potential?

What is the order of first,second,and third ionisation enthalpy? Why?

The order of ionisation potential between $He^{+}$ ion and $H$ atom (both species are in gaseous state) is

If first ionization enthalpies of element $X$ and $Y$ are $419 \ kJ \ mol^{-1}$ and $590 \ kJ \ mol^{-1}$,respectively and second ionization enthalpies of $X$ and $Y$ are $3069 \ kJ \ mol^{-1}$ and $1145 \ kJ \ mol^{-1}$,respectively. Then the correct statement is :-

The ionization potential $(IP)$ values for an element are given below. Identify the element:
$(IP)_1$ $7.1 \ eV$
$(IP)_2$ $10.3 \ eV$
$(IP)_3$ $36.2 \ eV$
$(IP)_4$ $39.1 \ eV$

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