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4 months ago

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What stress will shift the following equilibrium system to the right? 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔH= –98.8 kJ/mol Decreasing con

centration of SO2 Decreasing temperature Increasing concentration of SO3 Increasing volume

1 answer:

Alekssandra [3K]4 months ago

3 0

Answer:

Lowering the temperature.

Explanation:

According to Le Châtelier's principle, when an equilibrium is disturbed by an outside influence, the system reacts by shifting in the direction that counteracts that disturbance and restores balance.
Let's analyze the given options:

1) Reducing the concentration of SO₂,

A decrease in SO₂ concentration causes the reaction to move leftward to compensate for this drop.

2) Lowering the temperature,

Given ΔH = –98.8 kJ/mol, which implies the reaction releases heat and is exothermic.

Dropping the temperature means there's less heat, prompting the equilibrium to shift right to produce more heat and offset the change.

This is the correct option.

3) Increasing SO₃ concentration,

An increase in SO₃ shifts the reaction left to reduce the SO₃ levels.

4) Raising the volume,

Increasing volume lowers the pressure, causing the equilibrium to shift towards the side with more gas moles—that is, the left side.

Therefore, decreasing the temperature is the correct choice.

You might be interested in

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

2 months ago

What is another name for the molecular orbital theory of bonding in metals?

Alekssandra [3086]

The correct option is A. The band theory of metal explains how metals carry electricity by utilizing the electrons in their outer shells. When atomic orbitals of metals with similar energy levels merge, they create molecular orbitals and form bands. These bands facilitate the movement of electrons within metals, enabling them to conduct electricity.

4 0

2 months ago

The volume of a gas at 6.0 atm is 2.5 L. What is the volume of the gas at 7.5 atm at the same temperature?

castortr0y [3046]

Greetings!

The result is:

The new volume is: $2L$

Rationale:

Because the temperature remains constant, we can apply Boyle's Law to solve this issue.

Boyle's Law stipulates that:

$P_{1}V_{1}=P_{2}V_{2}$

Where,

P is the gas's pressure.

V is the gas's volume.

According to the information provided:

$V_{1}=2.5L\\P_{1}=6.0atm\\P_{2}=7.5atm$

Let's put the values into the equation:

$2.5L*6.0atm=7.5atm*V_{2}$

$2.5L*6.0atm=7.5atm*V_{2}\\\\V_{2}=\frac{2.5L*6.0atm}{7.5atm}=\frac{15L.atm}{7.5atm}=2L$

Consequently, the new volume is: $2L$

Wishing you a lovely day!

7 0

3 months ago