37.8 ~g ~N₂ O₅$37.8 \mathrm{~g} \mathrm{~N}_{2} \mathrm{O}_{5}$ was taken in a 1 ~L$1 \mathrm{~L}$ reaction vessel and allowed to undergo the following reaction at 500 ~K$500 \mathrm{~K}$: 2 N _ 2 O _ 5 (g) leftharpoons 2 N _ 2 O _ 4 (g) + O _ 2 (g)$$2 \mathrm {N} _ {2} \mathrm {O} _ {5 (\mathrm {g})} \rightleftharpoons 2 \mathrm {N} _ {2} \mathrm {O} _ {4 (\mathrm {g})} + \mathrm {O} _ {2 (\mathrm {g})}$$ The total pressure at equilibrium was found to be 18.65 bar. Then, Kp = _ _ _ _ × 10⁻²$\mathrm{Kp} = \_ \_ \_ \_ \times 10^{-2}$ [nearest integer] Assume N₂O₅$\mathrm{N}_2\mathrm{O}_5$ to behave ideally under these conditions Given: R = 0.082$\mathrm{R} = 0.082$ bar L mol⁻¹ K⁻¹$\mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$

Related Formula

Core Logic

Let us write down the dissociation dynamics for the reaction starting with a$a$ moles of A(g)$\mathrm{A(g)}$:

Step 1: Calculate Partial Pressures

The mole fractions (Xᵢ$X_i$) are:

• XA = (1-α)/(1+α)$X_A = \frac{1-\alpha}{1+\alpha}$
• XB = (α)/(1+α)$X_B = \frac{\alpha}{1+\alpha}$
• XC = (α)/(1+α)$X_C = \frac{\alpha}{1+\alpha}$

If the total pressure of the gas mixture at equilibrium is p$p$, the partial pressures are:

• pA = ( (1-α)/(1+α) ) p$p_A = \left( \frac{1-\alpha}{1+\alpha} \right) p$
• pB = ( (α)/(1+α) ) p$p_B = \left( \frac{\alpha}{1+\alpha} \right) p$
• pC = ( (α)/(1+α) ) p$p_C = \left( \frac{\alpha}{1+\alpha} \right) p$

Step 2: Relate K_p to alpha and p

Using the expression for Kₚ$K_p$:

Kₚ = (pB · pC)/(pA) = (((α)/(1+α))p · ((α)/(1+α))p)/(((1-α)/(1+α))p)$K_p = \frac{p_B \cdot p_C}{p_A} = \frac{\left(\frac{\alpha}{1+\alpha}\right)p \cdot \left(\frac{\alpha}{1+\alpha}\right)p}{\left(\frac{1-\alpha}{1+\alpha}\right)p}$

Since Kₚ$K_p$ is strictly a function of temperature, it remains constant.

Therefore, if total pressure p$p$ increases, the term (Kₚ)/(p)$\frac{K_p}{p}$ decreases, which demands that the term (α²)/(1-α²)$\frac{\alpha^2}{1-\alpha^2}$ must decrease. This is only possible if the degree of dissociation α$\alpha$ decreases.

Pattern Recognition

Le Chatelier's Principle Shortcut: For reactions with Δ ng > 0$\Delta n_g > 0$, raising the pressure pushes the system in the reverse direction to decrease the gas moles, which logically decreases the degree of dissociation α$\alpha$.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

Let us state Haber's process reaction:

This synthesis reaction is thermodynamically characterized by:

• Δ H° < 0$\Delta H^{\circ} < 0$ (Exothermic reaction)
• Δ S° < 0$\Delta S^{\circ} < 0$ (Decrease in the number of gaseous molecules, from 4 moles to 2 moles)

Both Δ H°$\Delta H^{\circ}$ and Δ S°$\Delta S^{\circ}$ remain relatively constant across the temperature range of interest.

N₂(g) + 3H₂(g) arrow 2NH₃$\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightarrow 2\mathrm{NH}_3$

^° = -ve$\Delta\mathrm{H}^\circ = -\text{ve}$

^° = -ve$\Delta\mathrm{S}^\circ = -\text{ve}$

(As gaseous moles decreases).

(1) As temperature increases - ^°RT$\frac{-\Delta\mathrm{H}^\circ_{\mathrm{R}}}{\mathrm{T}}$ , decreases

(2) ^° = -RT ln Keq$\Delta\mathrm{G}^\circ = -\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}}$

R ln Keq = - ^°T$\mathrm{R} \ln \mathrm{K}_{\mathrm{eq}} = -\frac{\Delta\mathrm{G}^\circ}{\mathrm{T}}$

(on increasing temperature in exothermic reaction Keq$\mathrm{K}_{\mathrm{eq}}$ decreases)

^°$\Delta\mathrm{H}^\circ$ and ^°$\Delta\mathrm{S}^\circ$ are almost constant with temperature.

This perfectly matches Graph (1).

Pattern Recognition

Since Haber's process is exothermic, Keq$K_{\text{eq}}$ must decrease as temperature increases (according to Le Chatelier's Principle). Because R ln Keq = -Δ G°/T$R \ln K_{\text{eq}} = -\Delta G^{\circ}/T$, the quantity -Δ G°/T$-\Delta G^{\circ}/T$ must also decrease with temperature.

Chapter Mix

Class 11 Chemistry: Equilibrium Class 11 Chemistry: Chemical Thermodynamics

Related Formula

Ideal Gas equation layout for aggregate systems:

Partial Pressure expression using mole fractions:

Core Logic

Let's tabulate equilibrium progress row-by-row:

• Reaction matrix:

• Given x = 0.04~mol$x = 0.04\mathrm{~mol}$ at equilibrium:
• nCO = 0.1 - 0.04 = 0.06~mol$n_{\mathrm{CO}} = 0.1 - 0.04 = 0.06\mathrm{~mol}$
• nCH₃OH = 0.04~mol$n_{\mathrm{CH_3OH}} = 0.04\mathrm{~mol}$
• Determine total moles via system pressure (P = 5~bar$P = 5\mathrm{~bar}$, V = 2~L$V = 2\mathrm{~L}$, T = 500K$T = 500\mathrm{K}$):

• Find remaining unknown hydrogen moles:

Step 1: Calculate Kp

Compute partial pressures using fractional allocation fractions (ntotal = 0.25$n_{\text{total}} = 0.25$):

• pCH₃OH = (0.04)/(0.25) × 5 = 0.8~bar$p_{\mathrm{CH_3OH}} = \frac{0.04}{0.25} \times 5 = 0.8\mathrm{~bar}$
• pCO = (0.06)/(0.25) × 5 = 1.2~bar$p_{\mathrm{CO}} = \frac{0.06}{0.25} \times 5 = 1.2\mathrm{~bar}$
• pH₂ = (0.15)/(0.25) × 5 = 3.0~bar$p_{\mathrm{H_2}} = \frac{0.15}{0.25} \times 5 = 3.0\mathrm{~bar}$

Substitute these pressures into the equilibrium expression:

Rounding to the nearest integer yields 74.

Pattern Recognition

Finding the total moles using the Ideal Gas Law from the final equilibrium pressure and volume cuts down steps, as it avoids explicitly computing the initial hydrogen amount 'a$a$' first.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Condition for precipitation step to happen dynamically:

where Ionic Product Qₛₚ = [A²⁺][Y^-]²$Q_{\mathrm{sp}} = [\mathrm{A}^{2+}][\mathrm{Y}^-]^2$.

Core Logic

When equal volumes are combined, total fluid volume doubles, meaning individual concentrations are exactly cut in half:

Let's calculate Qₛₚ$Q_{\mathrm{sp}}$ value for item layout (3):

• [A²⁺] = 2.0 × 10⁻²2 = 10⁻² M$[\mathrm{A}^{2+}] = \frac{2.0 \times 10^{-2}}{2} = 10^{-2}\ \mathrm{M}$
• [Y^-] = 2.0 × 10⁻²2 = 10⁻² M$[\mathrm{Y}^-] = \frac{2.0 \times 10^{-2}}{2} = 10^{-2}\ \mathrm{M}$
• Evaluating total value:

• Comparing arrays: 10⁻⁶ > 5.2 × 10⁻⁷$10^{-6} > 5.2 \times 10^{-7}$, confirming precipitation conditions are satisfied.

Pattern Recognition

Do not skip the dilution factor! Halving initial chemical molar values before calculating the reaction quotients prevents incorrect combinations.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

For the initial solution:

When we dilute the solution by adding an equal volume of water, the final volume is doubled (V₂ = 2V₁$V_2 = 2V_1$). Thus, the final concentration is halved:

Now, calculate the new pH$\text{pH}$:

Pattern Recognition

Diluting any strong acid by 2$2$ times increases the pH$\text{pH}$ by exactly ₁₀ 2 ≈ 0.30$\log_{10} 2 \approx 0.30$. Thus, 1.0 + 0.3 = 1.3$1.0 + 0.3 = 1.3$ immediately.

Chapter Mix

Class 11 Chemistry: Ionic Equilibrium