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3 years ago

According to Arrhenius, NH4+ an acid or a base? Write an equation to support

Chemistry

1 answer:

sergejj [24]3 years ago

8 0

According to Arrhenius definition of acids, $NH4^+$ is an acid.

According to Arrhenius definition of acids and bases, acid is any substance that produces hydrogen ion in solution as its only positive ion.

Following this definition, let us now consider what happens when $NH4^+$ is introduced into a water;

$NH4^+$ (aq)-------> NH3(aq) + $H^+$ (aq)

Hence, according to Arrhenius definition of acids, $NH4^+$ is an acid.

For another example of Arrhenius acids, see

brainly.com/question/516295

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Draw an energy diagram for an endothermic reaction without a catalyst (use a solid line) and with a catalyst (use a dotted line)

olchik [2.2K]

- Endothermic reaction : so the energy of product is higher than the energy of reactant.
- Catalyst lowers the activation energy so it increase the rate of reaction.
- The graph is attached:

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3 years ago

Read 2 more answers

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

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

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

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

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

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3 years ago

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3 years ago

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3 years ago