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1 year ago

Consider each possible structure of carbon dioxide, c o 2. Determine whether the structure is correct and justify your decision.

Chemistry

1 answer:

Mrrafil [7]1 year ago

6 0

This structure is correct due to the total number of bonds and electrons

We employ the octet rule, which states that all atoms in a compound are expected to follow, to check the accuracy of any chemical structure. The octet rule is precisely satisfied for each atom in the depicted structure of carbon IV oxide. The valence shell of each atom in the molecule is surrounded by eight electrons. We may thus infer that this structure is accurate given the total number of bonds and electrons since CO2 has sixteen valence electrons.

It possesses a negative charge of 1.602176634 coulombs, the basic unit of electric charge, which is comparable to its negative charge. The rest mass of an electron is 9.1093837015 10^-31 kg, or just 1/1,836 the mass of a proton. An electron is therefore considered to have practically negligible mass in comparison to a proton or neutron, and its matter is not taken into consideration when calculating an atom's mass number.

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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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Peer review involves subjecting the author's scholarly work and research to the scrutiny of other experts in the same field to check its validity and evaluate its suitability for publication. A peer review helps the publisher decide whether a work should be accepted.

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<h3>What is decay constant unit?</h3>

Definition. The decay constant (symbol: λ and units: s−1 or a−1) of a radioactive nuclide is its probability of decay per unit time. The number of parent nuclides P therefore decreases with time t as dP/P dt = −λ. The energies involved in the binding of protons and neutrons by the nuclear forces are ca.

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