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3 years ago

Because amines are basic, they can often accept a proton. draw the protonated structure of n-propylamine

1 answer:

3 0

Answer is: CH₃-CH₂-CH₂-NH₃⁺.
Amine group (-NH₂) has neutral charge, bet when accepts one proton (H⁺) has positive charge.
Amines are compounds that contain a basic nitrogen atom with a lone pair of electrons. Amines are derivatives of ammonia, where is one or more hydrogen atoms replaced by a substituent such as an alkyl or aryl group (in this compound propyl group).

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Answer:

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3 years ago

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a

igor_vitrenko [27]

Answer: The equilibrium constant for this reaction is $1.068\times 10^{6}$

Explanation:

The equation used to calculate standard Gibbs free change is of a reaction is:

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

For the given chemical reaction:

$Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)$

The equation for the standard Gibbs free change of the above reaction is:

$\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]$

We are given:

$\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol$

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = $58^oC=[273+58]K=331K$

$K_{eq}$ = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

$-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}$

Hence, the equilibrium constant for this reaction is $1.068\times 10^{6}$

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3 years ago

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Viefleur [7K]

c.
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