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Standard XII

Chemistry

Heat of Reaction

An element un...

Question

An element undergoes a reaction as shown: $X + 2 e^{-} \to X^{2 -}$ The energy released is $30.87 eV/atom$ . This energy is used to dissociate $4 gm$ of hydrogen molecules equally into $H^{+}$ and $H^{}$ (where $H^{}$ is the excited state of H atoms). The electron travels in an orbit whose circumference is equal to four times its de Broglie’s wavelength, in $H^{*}$ . Determine the least number of moles of $X$ that would be required for this to happen. Given: I.E. of $H$ $= 13.6 eV/atom$ , bond energy of $H_{2} = 4.526 ev/molecule$

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Solution

The correct option is B 2
$2 π r = 4 λ; n = 4$
Total energy required + total energy released = 0
$2 \times 4.526 e V \times N_{A} + 2 \times 13.6 \times N_{A} + 2 \times 13.6 \times (1 - \frac{1}{16}) \times N_{A} - 30.87 \times x \times N_{A} = 0$
$x = 2$
Hence, the moles of $X$ required= 2

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Similar questions

Q. An element undergoes a reaction as shown:

X + 2 e^{-} \to X^{2 -}

, energy released

= 30.87

eV/atom. If the energy released, is used to dissociate

4

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H_{2}

molecules, equally into

H^{+}

and

H^{*}

, where

H^{*}

is excited state of H atoms where the electron travels in orbit whose circumference equal to four times ts de Broglie's wavelength. Determine the least moles of

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Given: I.E. of

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eV/atom, bond energy of

H_{2} = 4.526

eV/ molecule.

Q. A hypothetical element X undergoes a reaction as shown:

X + 2 e^{-} \to X^{2 -}

, energy released = 30.87 eV/atom. If the energy released is used to dissociate 4 g of

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H^{+}

and

H^{*}

, where

H^{*}

is the excited state of H atoms where the electron travels in orbit whose circumference is equal to four times its de-broglie's wavelength. Determine the least moles of X that would be required:
Given: I.E of H = 13.6 eV/atom , bond energy of

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= 4.526 eV/molecule.

Q. An element undergoes reaction as shown:

X + 2 e -

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= 30.87 e V / a t o m

. If the energy released, is used to dissociate

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g of

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molecules, equally into

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and

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, where

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is excited state of H atoms, where the electrons travels in in orbit whose circumference equal to four times it's de Broglie's wavelength. Determine the least moles of X that would be required:
Given : I.E. of H =

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Q. What is the energy required to move the electron from the ground state of H atom to the first excited state? Given that the ground state energy of H atom is 13.6 eV and that the energy E

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The correct option is B 22πr=4λ; n=4 Total energy required + total energy released = 0 2×4.526 eV×NA+2×13.6×NA+2×13.6×(1−116)×NA−30.87×x×NA=0 x=2 Hence, the moles of X required= 2

The correct option is B 2
$2 π r = 4 λ; n = 4$
Total energy required + total energy released = 0
$2 \times 4.526 e V \times N_{A} + 2 \times 13.6 \times N_{A} + 2 \times 13.6 \times (1 - \frac{1}{16}) \times N_{A} - 30.87 \times x \times N_{A} = 0$
$x = 2$
Hence, the moles of $X$ required= 2