Law Of Chemical Equilibrium And Equilibrium Constant

Given:

• Reaction: X(s) ⇌ Y(s) + Z(g)
• Plot of ln(p_Z/p⁰) vs 10⁴/T is linear.
• From the graph:
  • At (10, -3)
  • At (12, -7)
• Gas constant: R = 8.314 J mol^-1 K^-1
• Standard enthalpy relation:
  d(ln K)/d(1/T) = -ΔH⁰/R

Step 1: Determine the slope of the line

We use the two given points (10, -3) and (12, -7) from the graph to find the slope (m):

Slope = (y₂ - y₁) / (x₂ - x₁) = (-7 - (-3)) / (12 - 10) = (-4)/2 = -2

Step 2: Use the Van’t Hoff equation

From the Van’t Hoff equation:
Slope = -ΔH⁰ / R
Therefore,
-2 = -ΔH⁰ / R

Step 3: Solve for ΔH⁰

Multiply both sides by R = 8.314 J mol^-1 K^-1:
ΔH⁰ = 2 × 8.314 × 10⁴ (Note: the x-axis is in 10⁴/T units)
ΔH⁰ = 166280 J mol^-1
ΔH⁰ = 166.28 kJ mol^-1

Final Answer: 166.28 kJ mol^-1

The equation that relates the standard Gibbs free energy change ( ), the enthalpy change ( ), and the entropy change ( ) is:

where T is the temperature in kelvin and R is the gas constant.
The equation that relates the equilibrium constant (K) and the standard Gibbs free energy change is:

We can combine these two equations to eliminate ΔG° and solve for K.
Given that and P° = 1 bar

The slope of the graph in the image is equal to .

Slope =

We can use the gas constant value (R = 8.314 J/mol*K) to find ΔH°.

We can also use the graph to find the temperature (T) at which the pressure is 10^7. The temperature is 10^3 K.

We can substitute the values of ΔH° and T into equation (iii) to solve for ΔS°.

So, the correct answer is

Le Chatelier's Principle and Equilibrium Reaction

The correct answer is 8.

The given graph depicts the progress of product formation for the reaction at two different temperatures, and . The correct answer is 8, which refers to the ratio of to .

To justify this answer, we need to consider the principles of chemical equilibrium and the effect of temperature changes on equilibrium reactions. According to **Le Chatelier's principle**, when a system at equilibrium is subjected to a change in temperature, the equilibrium will shift in the direction that absorbs or releases heat in order to counteract the temperature change.

In the graph provided, it's evident that the equilibrium position at temperature favors the product formation (P) more than at temperature . This suggests that increasing the temperature from to has shifted the equilibrium towards the products. This is in line with Le Chatelier's principle because the forward reaction is endothermic, absorbing heat. Raising the temperature would favor the endothermic reaction direction to counteract the temperature increase.

As a result, the ratio of the equilibrium constants at the two temperatures is given by:

Final Answer

The correct value of the ratio is .

Gibbs Free Energy Calculation

Given: The chemical equilibrium for the reaction:

Initially: 6 moles of A and 0 moles of P.

At equilibrium: moles of A and moles of P.

Solution:

At temperature , we are given . Therefore, the equilibrium constant is:

At temperature , we are given . Therefore, the equilibrium constant is:

Using the equation for Gibbs free energy change:

Now, we subtract the two equations to find the difference in Gibbs free energy:

Therefore, , and since , we conclude that:

The value of is given by where .

The correct answer is .

Step 1: Write the dissociation equation of barium iodate ( ) in water: Step 2: Given, - Volume and concentration of barium nitrate , - Volume and concentration of sodium iodate , Step 3: Calculate the total volume after mixing: Step 4: Calculate initial concentrations of ions from mixing before any precipitation: Step 5: Let the solubility of barium iodate in the mixed solution be . At equilibrium: Since is very small, will be very small compared to initial concentrations, so we approximate: Step 6: Calculate the ion product (Q) to check if precipitation occurs: Since , the solution is supersaturated and precipitation will occur. Step 7: At equilibrium, Since is small compared to initial concentrations, use approximation: This contradiction implies we cannot neglect . Step 8: Define: Write the equilibrium expression: Since initial concentrations are much larger than , and is very small, precipitation will reduce ion concentrations to the saturation level. Step 9: For calculation of solubility , we consider only the amount of barium iodate that dissolves, ignoring initial concentrations. Solubility of barium iodate in pure water: Step 10: In presence of common ions, the solubility decreases significantly. Approximate the effect by calculating the concentration from initial ions and use: Given that is very small, the solubility can be approximated by solving: But since , practically no more barium iodate will dissolve, so solubility . Step 11: Therefore, solubility .
Final Answer: