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Sergeu [11.5K]
3 years ago
11

What type of chemical bond forms between positive and negative ions? (1 point)?

Chemistry
2 answers:
Rainbow [258]3 years ago
8 0

Ionic is the correct answer

vlada-n [284]3 years ago
7 0

Ionic bond is formed between positive and negative ions.

Positive ion is known as cation (electropositive metal)

Negative ion is known as anion ( electronegative non-metal)

The bond formed between a metal (positive ion) and non-metal (negative ions) is known as Ionic bond.

Ionic form is formed when one or more electrons from the valence shell of an atom (electropositive metal ) are completely transferred to the valence shell of another atom (electronegative non-metal). Ionic bonding is the complete transfer of valence electron(s) between oppositely charged atoms.

For example: Bond between NaCl(sodium chloride) is an ionic bond in which Na (sodium ) is positive ion(metal) which have +1 charge and Cl (chlorine) is negative ion(non-metal) which have -1 charge and the bond formed between Na+ and Cl- is ionic bond.

Na^{+}(positive ion) + Cl^{-}(negative ion)\rightarrow Ionic bond

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3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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