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Byju's Answer

Standard XII

Chemistry

Gay Lussac's Law, Avagadro's Law

If u are give...

Question

If u are given Avogadro number (N) of atoms of a gas X if half of atoms are converted into X(g) + by energy $Δ$ H. the IE of X is?

2 Δ H / N

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2 N / Δ H

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Δ H / 2 N

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N / Δ H

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Solution

The correct option is A $2 Δ H / N$
$N$ =Avogadro number= $6.023 \times 10^{23}$
Each atom gives one electron out. Among half of $N$ atoms are ionized.
Ionization energy of element $X = \frac{Δ H}{\frac{N}{2}}$
$∴ \frac{2 Δ H}{N}$ .

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Similar questions

Q. If you are given Avogadro's number of atoms of a gas

X

. If half of the atoms are converted into

X_{(g)}^{+}

by energy

Δ H

. The IE of

X

is :

Q. N

_{0}

/2 atoms of X(g) are converted into X

^{+}

(g) by energy E

_{1}

. N

_{0}

/2 atoms of X(g) are converted into X

^{-}

(g) by energy E

_{2}

. Hence, ionisation potential and electron affinity of X(g) are:

If bond energy of $N - H, H - H$ and $N \equiv N$ are a, b and c respectively, then $Δ H$ for the reaction, $2 N H_{3} \to N_{2} + 3 H_{2}$ is?

\frac{N_{0}}{2}

atoms of

X (g)

are converted into

X^{+} (g)

by absorbing

E_{1}

energy.

2 N_{0}

atoms of

X (g)

are converted into

X^{-} (g)

by releasing

E_{2}

energy. Calculate the ionisation enthalpy and electron gain enthalpy of

X (g)

per atom.

\frac{N_{0}}{2}

atoms of

X (g)

are converted into

X^{+} (g)

by absorbing

E_{1}

energy.

2 N_{0}

atoms of

X (g)

are converted into

X^{-} (g)

by releasing

E_{2}

energy. Calculate the ionisation enthalpy and electron gain enthalpy of

X (g)

per atom.

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If u are given Avogadro number (N) of atoms of a gas X if half of atoms are converted into X(g) + by energy ΔH. the IE of X is?

The correct option is A 2ΔH/NN=Avogadro number= 6.023×1023Each atom gives one electron out. Among half of N atoms are ionized.Ionization energy of element X=ΔHN2∴2ΔHN .

If u are given Avogadro number (N) of atoms of a gas X if half of atoms are converted into X(g) + by energy $Δ$ H. the IE of X is?

The correct option is A $2 Δ H / N$
$N$ =Avogadro number= $6.023 \times 10^{23}$
Each atom gives one electron out. Among half of $N$ atoms are ionized.
Ionization energy of element $X = \frac{Δ H}{\frac{N}{2}}$
$∴ \frac{2 Δ H}{N}$ .