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3 years ago

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The ionization energy is the energy needed to remove an electron from an atom. In the bohr model of the hydrogen atom, this mean

s exciting the electron to the n = ∞ state. What is the ionization energy in kj/mol, for hydrogen atoms initially in the n = 2 energy level?

1 answer:

Salsk061 [2.6K]3 years ago

6 0

The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.

The <em>first ionization energy</em> of hydrogen is 1312.0 kJ·mol⁻¹.

Thus, H atoms in the <em>n</em> = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when <em>n</em> = ∞.

According to Bohr, Eₙ = k/<em>n</em>².

If <em>n</em> = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.

If <em>n</em> = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹

∴ The ionization energy from <em>n</em> = 2 is 328 kJ·mol⁻¹ .

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a

Mandarinka [93]

There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

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where pKa = - log [Ka]
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Given:
Ka = 1.8x10^-5
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NaC2H3O2 = 0.1 mol

Solution:
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