HBrO (aq) + H2O (l) ⇋ H3O+ (aq) + BrO- (aq)

A sample of solid naphthalene is introduced into an evacuated flask. Use the data below to calculate the equilibrium vapor press

Tems11 [2777]

Answer: The vapor pressure of naphthalene within the flask remains at $2.906\times 10^{-4}$ atm.

Explanation:

The transformation from solid naphthalene to its gaseous form follows the equilibrium reaction:

$C_{10}H_8(s)\rightleftharpoons C_{10}H_8(g)$

The formula employed to determine the enthalpy change for the reaction is:

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

The formula for calculating the enthalpy change regarding the aforementioned reaction is:

$\Delta H^o_{rxn}=(1\times \Delta H^o_f_{(C_{10}H_8(g))})-(1\times \Delta H^o_f_{(C_{10}H_8(s))})$

The provided information includes:

$\Delta H^o_f_{(C_{10}H_8(s))}=78.5kJ/mol\\\Delta H^o_f_{(C_{10}H_8(g))}=150.6kJ/mol$

Substituting the values into the previous equation produces:

$\Delta H^o_{rxn}=(1\times 150.6)-(1\times 78.5)=72.1kJ/mol$

The formula utilized to compute Gibbs free energy change is of a reaction:

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

The equation for the enthalpy change for the reaction is:

$\Delta G^o_{rxn}=(1\times \Delta G^o_f_{(C_{10}H_8(g))})-(1\times \Delta G^o_f_{(C_{10}H_8(s))})$

The given factors include:

$\Delta G^o_f_{(C_{10}H_8(s))}=201.6kJ/mol\\\Delta G^o_f_{(C_{10}H_8(g))}=224.1kJ/mol$

By inserting values from the above equation, we arrive at:

$\Delta G^o_{rxn}=(1\times 224.1)-(1\times 201.6)=22.5kJ/mol$

For the calculation of $K_1$ (at 25°C) regarding the provided value of Gibbs free energy, the following relationship is applied:

$\Delta G^o=-RT\ln K_1$

where,

$\Delta G^o$ = Gibbs free energy = 22.5 kJ/mol = 22500 J/mol (Conversion factor: 1kJ = 1000J)

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

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

$K_1$ = equilibrium constant at 25°C =?

Inserting values into the above equation yields:

$22500J/mol=-(8.314J/Kmol)\times 298K\times \ln K_1\\\\K_1=1.14\times 10^{-4}$

To determine the equilibrium constant at 35°C, we refer to the equation proposed by Arrhenius, which states:

$\ln(\frac{K_2}{K_1})=\frac{\Delta H}{T}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$K_2$ = Equilibrium constant at 35°C =?

$K_1$ = Equilibrium constant at 25°C = $1.14\times 10^{-4}$

$\Delta H$ = Enthalpy change of the reaction = 72.1 kJ/mol = 72100 J

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

$T_1$ = Initial temperature = $25^oC=[273+25]K=298K$

$T_2$ = Final temperature = $35^oC=[273+35]K=308K$

By plugging values into the equation above, we obtain:

$\ln(\frac{K_2}{1.14\times 10^{-4}})=\frac{72100J/mol}{8.314J/K.mol}(\frac{1}{298}-\frac{1}{308})\\\\K_2=2.906\times 10^{-4}$

In order to calculate the partial pressure of naphthalene at 35°C, we utilize the equation for $K_p$ , which is:

$K_p=\frac{p_{C_{10}H_8(g)}}{p_{C_{10}H_8(g)}}=p_{C_{10}H_8(g)$

The partial pressure of the solid phase is considered to be 1 at equilibrium.

Therefore, the value for $K_2$ will equal $K_p$

$p_{C_{10}H_8}=2.906\times 10^{-4}$

Consequently, the partial pressure of naphthalene at 35°C is $2.906\times 10^{-4}$ atm.

3 0

1 month ago

Which of the following reactions is a redox reaction? (a) K2CrO4 + BaCl2 → BaCrO4 + 2KCl (b) Pb2+ + 2Br- → PbBr2 (c) Cu + S → Cu

castortr0y [3046]

(c) Cu + S → CuS is classified as a redox reaction

Explanation:

The following reactions are presented:

(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl

(b) Pb²⁺ + 2 Br⁻ → PbBr₂

(c) Cu + S → CuS

Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:

Cu + S → CuS

In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.

Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).

Learn more about:

redox reactions

7 0

2 months ago

BRAINLIESTTT ASAP!!!

lorasvet [2795]

Although I may not be the smartest, I can definitely answer.

This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.

8 0

3 months ago

Read 2 more answers

A sealed 10.0 L flask at 400 K contains equimolar amounts of ethane and propanol in gaseous form. Which of the following stateme

castortr0y [3046]

Answer:

(C) The average speed of molecules in ethane is the same as that of propanol.

Explanation:

In gas behavior, temperature is directly linked with speed. At a constant temperature, speed remains consistent. Also, we understand that ideal gases exhibit uniform behavior, irrespective of their type.

8 0

2 months ago

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Water's surface tension and heat storage capacity are accounted for by its ______. A) orbitals B) weight C) hydrogen bonds D) ma

lorasvet [2795]

The answer is C: hydrogen bonds. Explanation: The surface tension and the capacity for heat storage in water are due to its hydrogen bonds. Water molecules have a strong attraction to one another through hydrogen bonding. These bonds are continuously forming and breaking within water molecules. The result of this hydrogen bonding is surface tension, which allows water to have a greater capacity for heat retention. Consequently, during the night, the temperature on Earth drops much faster than it does for water, as water gradually releases heat, helping maintain a moderate atmospheric temperature at night.

5 0

2 months ago