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UNO [17]
3 years ago
8

Which is the correct statement regarding the relative Rf values of the starting methyl benzoate vs the product, methyl m-nitrobe

nzoate on a silica gel TLC plate. The product has a higher Rf value on a silica gel TLC plate because it is less polar than the starting methyl benzoate. The product has a lower Rf value on a silica gel TLC plate because it is less polar than the starting methyl benzoate. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate. The product has a lower Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate. 1.5 points QUESTION 2 Methyl benzoate is more reactive than benzene in Electrophilic Aromatic Substition [EAS] reactions. True False 1 points QUESTION 3 Methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition [EAS] reactions. True False
Chemistry
1 answer:
Marianna [84]3 years ago
5 0

Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

Explanation:

In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.

In comparing the electrophillic aromatic substitution of m-nitrobenzoate  and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

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Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta
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<u>Answer:</u> The \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)

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

\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]

We are given:

\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ

Putting values in above equation, we get:

-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ

Hence, the \Delta H_f for HCN (g) in the reaction is 135.1 kJ/mol.

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