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Question

Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

For the reaction, X(s)Y(s)+Z(g)X (s) \rightleftharpoons Y (s)+ Z (g), the plot of lnpZp0\ln \frac{p_{ Z }}{p^{0}} versus 104T\frac{10^{4}}{T} is given below (in solid line), where pzp_{ z } is the pressure (in bar) of the gas ZZ at temperature TT and p0=1p^0=1 bar
For the reaction, 𝐗𝑠 ⇌ 𝐘𝑠 + 𝐙𝑔, the plot of ln 𝑝𝐙 𝑝o versus 104 𝑇 is given below in solid line, where 𝑝𝐙 is the pressure in bar of the gas Z at temperature T and 𝑝 o = 1 bar.
(Given, d(lnK)d(1T)=ΔH0R\frac{ d (\ln K)}{ d \left(\frac{1}{T}\right)}=-\frac{\Delta H^{0}}{R}, where the equilibrium constant, K=pzp0K=\frac{p_{z}}{p^{0}} and the gas constant, R=8314JK1mol1R=8314 \,J \,K ^{-1} mol ^{-1} )
The value of standard enthalpy, ΔH0\Delta H^{0} (in kJmol1kJ\, mol ^{-1} ) for the given reaction is ______

Answer

The value of standard enthalpy, ΔH0\Delta H^{0} (in kJmol1kJ\, mol ^{-1} ) for the given reaction is 166.28.

Explanation

Solution

The value of standard enthalpy, ΔH0\Delta H^{0} (in kJmol1kJ\, mol ^{-1} ) for the given reaction is 166.28.