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zmey [24]
3 years ago
5

A certain electromagnetic wave a wavelength of 415 nm. What is the frequency of the wave in hz?

Chemistry
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

The frequency of the electromagnetic wave is 7.22891566 × 10¹⁴ Hz

Explanation:

The wavelength of the electromagnetic wave, λ = 415 nm

The speed of an electromagnetic wave, c ≈ 3.0 × 10⁸ m/s

Given that an electromagnetic wave is a periodic wave, we have;

The speed of the electromagnetic wave, c = f×λ

Where;

f = The frequency of the electromagnetic wave

Therefore, we have;

f = c/λ

From which we have;

f = (3.0 × 10⁸ m/s)/(415 nm) = 7.22891566 × 10¹⁴ /s = 7.22891566 × 10¹⁴ Hz

The frequency of the electromagnetic wave, f = 7.22891566 × 10¹⁴ Hz

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<u>Answer:</u> The value of \Delta G^o of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

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  • The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]

We are given:

\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol

  • The equation used to calculate entropy change is of a reaction is:

\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]

We are given:

\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

where,

\Delta H^o_{rxn} = standard enthalpy change of the reaction =-67200 J/mol

\Delta S^o_{rxn} = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol

Hence, the value of \Delta G^o of the reaction is 28.38 kJ/mol

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