Question

For the reaction, $X (s) \rightleftharpoons Y (s)+ Z (g)$, the plot of $\ln \frac{p_{ Z }}{p^{0}}$ versus $\frac{10^{4}}{T}$ is given below (in solid line), where $p_{ z }$ is the pressure (in bar) of the gas $Z$ at temperature $T$ and $p^0=1$ bar

 (Given, $\frac{ d (\ln K)}{ d \left(\frac{1}{T}\right)}=-\frac{\Delta H^{0}}{R}$, where the equilibrium constant, $K=\frac{p_{z}}{p^{0}}$ and the gas constant, $R=8314 \,J \,K ^{-1} mol ^{-1}$ ) 
The value of standard enthalpy, $\Delta H^{0}$ (in $kJ\, mol ^{-1}$ ) for the given reaction is ______

Answer 1

Correct Answer: 166.28

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