1. Tardigrade
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  3. Chemistry
  4. For a hypothetical reaction, A(g) +B(g) leftharpoons C(g) +D(g) a graph between log K and T-1 is straight line as follows, where θ = tan-1 0.5 and OA = 10. Assuming ΔH° is independent of temperature, the equilibrium constants at 298 K and 798 K are respectively

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Q. For a hypothetical reaction, $A_{(g)} +B_{(g)} \rightleftharpoons C_{(g)} +D_{(g)}$ a graph between $log\, K$ and $T^{-1}$ is straight line as follows, where $θ = tan^{-1}\, 0.5$ and $OA = 10$.
Assuming $ΔH^°$ is independent of temperature, the equilibrium constants at $298\, K$ and $798\, K$ are respectivelyChemistry Question Image

Equilibrium

Solution:

Slope $=tan\,\theta=\frac{\Delta H^{°}}{2.303\,R}$
$0.5=\frac{\Delta H^{°}}{2.303\times8.314}$
$\Delta H^{°}=9.574\,J\,mol^{-1}$
Intercept $= log_{10}\, A = 10, A = 10^{10}$
$log\,K=10-\frac{9.574}{2.303\times8.314\times298}$
$K=9.96\times10^{9}$
As $ΔH^°$ is independent of temperature, equilibrium constant remains almost constant. It means even at $798\, K$, equilibrium constant $K = 9.98 × 10^9$