Question

For a hypothetical reaction, $A_{(g)}+B_{(g)}\rightleftharpoons C_{(g)}+D_{(g)}$ a graph between $logK$ and $T^{-1}$ is straight line as follows, where $\theta =ta{n}^{-1}0.5$ and $OA=10$.
Assuming $\Delta H^{°}$ is independent of temperature, the equilibrium constants at $298K$ and $798K$ are respectively

• A. $K_1=9.96\times 10^9,K_2=2.76\times 10^{10}$
• B. $K_1=3.79\times 10^8,K_2=9.98\times 10^9$
• C. $K_1=7.96\times 10^9,K_2=9.96\times 10^9$
• D. $K_1=9.98\times 10^9,K_2=9.96\times 10^9$

Answer 1

Answer: (D)

Slope $=tan\theta =\frac{\Delta H^{°}}{2.303R}$
$0.5=\frac{\Delta H^{°}}{2.303\times 8.314}$
$\Delta H^{°}=9.574Jmo{l}^{-1}$
Intercept $=lo{g}_{10}A=10,A=10^{10}$
$logK=10-\frac{9.574}{2.303\times 8.314\times 298}$
$K=9.96\times 10^9$
As $\Delta H^{°}$ is independent of temperature, equilibrium constant remains almost constant. It means even at $798K$, equilibrium constant $K=9.98\times 10^9$