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2 years ago

Calculate the value deltaG°

Chemistry

1 answer:

atroni [7]2 years ago

4 0

Answer:

ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.

Explanation:

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The oxidation state of phosphorus is +3 in

IgorC [24]

Answer:

b) Phosphorus acid

Explanation:

To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:

Orthophosphoric acid H₃PO₄

Phosphorus acid H₃PO₃

Metaphosphoric acid HPO₃

Phyrophosphoric acid H₄P₂O₇

Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:

Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.

H₃PO₃:

we know the oxidation state of H = +1

O = -2

The oxidation state of P is unknown. We can express this as an equation:

3(+1) + P + 3(-2) = 0

3 + P -6 = 0

P-3 = 0

P = +3

6 0

2 years ago

If the half-life of hydrogen-3 is 11.8 years, after two half-lives the radioactivity of a sample will be reduced to one-half of

maw [93]

Answer:

False

Explanation:

Half life is the time period at which the concentration of the radioactive substance in decay reduced to half.

<u>Thus, if the hydrogen-3 has gone 2 half lives, it means that it has first reduced to its half and then again the half of what it was, i.e. 1/4</u>

Thus, after two successive half-lives, the concentration must be 1/4 of the initial concentration and hence, the statement is false.

4 0

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 $(\Delta S^o)$ .

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

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