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 ______
(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 ______
Correct Answer: 166.28
Solution and Explanation
Given:
- Reaction: X(s) β Y(s) + Z(g)
- Plot of ln(pZ/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) = -ΞH0/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 Vanβt Hoff equation
From the Vanβt Hoff equation:
Slope = -ΞH0 / R
Therefore,
-2 = -ΞH0 / R
Step 3: Solve for ΞH0
Multiply both sides by R = 8.314 J mol-1 K-1:
ΞH0 = 2 Γ 8.314 Γ 104 (Note: the x-axis is in 10β΄/T units)
ΞH0 = 166280 J mol-1
ΞH0 = 166.28 kJ mol-1
Final Answer: 166.28 kJ mol-1
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(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, $R=8.314 \,J \,K ^{-1} mol ^{-1}$ )
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(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, $R=8314 \,J \,K ^{-1} mol ^{-1}$ )- JEE Advanced - 2021
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B β C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and KbΒ are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.