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3 years ago

Oxalic acid, H2C2O4, reacts

Chemistry

1 answer:

Dominik [7]3 years ago

6 0

Answer:0.005M

Explanation:

First deduce the oxidation and reduction half equations and from that obtain the balanced redox reaction equation. From that, the number of moles of reacting species are seen from the stoichiometry of the reaction from which the number of moles of oxalate is obtained and substituted to obtain the molar concentration of oxalate.

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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

3 0

3 years ago

. A large quantity of very dilute aqueous HCl solution is neutralized by the addition of the stoichiometric amount of a 10-mol-%

ad-work [718]

Answer:

Make the question more clear for me

Explanation:

5 0

3 years ago

Without consulting Appendix B, arrange each group in order of increasing standard molar entropy (S°). Explain.(c) SF₆(g), SF₄(g)

Andre45 [30]

The increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

<h3>What is Entropy? </h3>

Entropy is defined as the randomness of the particle. It depends on temperature and pressure or number of particle per unit volume.

It is directly proportional to the temperature and pressure of the gas.

<h3>What is Standard Molar Entropy? </h3>

The standard molar entropy is defined as the entropy content of the one mole of pure substance at the standard state of temperature and pressure of interest.

The standard molar entropy is also defined as the total amount of entropy which 1 mole of the substance acquire, as it is brought from 0K to standard conditions of temperature and pressure.

The standard molar entropy depends on the molas mass of atom, molecules or compound.

SF₄(g) has lower standard molar entropy. Due to less complexity of this molecules.

While, complexity increases from SF₆(g) to S₂F₁₀(g). Therefore, the standard molar entropy of S₂F₁₀(g) is greater than SF₆(g).

Thus, we concluded that the increasing order of standard molar entropy (S°) is as follow:

SF₄(g) < SF₆(g) < S₂F₁₀(g)

learn more about standard molar entropy:

brainly.com/question/15908262

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7 0

2 years ago

Which variable is tested during a lab investigation

Licemer1 [7]

Actually, it could indeed be the independent variable.

5 0

3 years ago

What is the pH of 0.12M HNO3

Phoenix [80]

Answer:

pH= 0.92

Explanation:

HNO3-> H^+ +NO3^-

HNO3 is a strong acid, so it fully dissociates

[HNO3] = 0.12M [H^+] = 0.12M

pH= -log[H^+]

pH=-log[.12] = 0.92

pH = 0.92

4 0

3 years ago