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alukav5142 [94]
4 years ago
14

it takes 463 kJ/mol to break an oxygen-hydrogen single bond. calculate the maximum wavelength of light for which an oxygen-hydro

gen single bond could be broken by absorbing a single photon
Chemistry
1 answer:
Marianna [84]4 years ago
6 0

Answer:

  • <u>2.59 × 10⁻⁷ m  = 259 nm</u>

Explanation:

You need to calculate the wavelength of a photon with an energy equal to 463 kJ/mol, which is the energy to break an oxygen-hydrogen atom.

The energy of a photon and its wavelength are related by the Planck - Einstein equation:

  • E = hν,

Where:

  • E = energy of a photon,
  • h = Planck constant (6.626 × 10⁻³⁴ J . s) and
  • ν = frequency of the photon.

And:

  • ν = c / λ

Where:

  • c = speed of light (3.00 × 10⁸ m/s in vacuum)
  • λ = wavelength of the photon

Thus, you can derive:

  • E = h c / λ

Solve for λ:

  • λ = h c / E

Before substituting the values, convert the energy, 463 kJ/ mol, to J/bond

  • 463 kJ/ mol × 1,000 J/kJ × 1 mol / 6.022 × 10 ²³ atom × 1 bond / atom

       = 7.69×10²³ J / bond

Substitute the values and use the energy of one bond:

  • λ = 6.626 × 10⁻³⁴ J . s × 3.00 × 10⁸ m/s / 7.69×10²³ J = 2.59 × 10⁻⁷ m

The wavelength of light is usually shown in nanometers:

  • 2.59 × 10⁻⁷ m × 10⁹ nm / m = 259 nm ← answer
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<u>Answer:</u>

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<u>For b:</u> The activity coefficient of potassium ions is 0.851

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To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}       ........(1)

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Calculating the ionic strength by using equation 2:

C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M

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Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{  (known)}\\\mu=0.01M

Putting values in equation 1, we get:

-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676

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  • <u>For b:</u>

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Calculating the ionic strength by using equation 2:

C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.025M

Putting values in equation 1, we get:

-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851

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Calculating the ionic strength by using equation 2:

C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2

Putting values in equation 2, we get:

\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M

Now, calculating the activity coefficient of K^{+} ion in the solution by using equation 1:

Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{  (known)}\\\mu=0.06M

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