TS EAMCET 9 May 2024 Shift 1 Solved Paper

The Arrhenius equation describes the temperature dependence of the rate constant $k$ for a reaction, and is given by:$k=A e^{-\;\frac{E_a}{RT}}$Taking the natural logarithm of both sides gives:$\ln k=\ln A-\;\frac{E_a}{R} \cdot \;\frac{1}{T}$This equation represents a linear plot of $\ln k$ vs. $\;\frac{1}{T}$, where the slope is $-\;\frac{E_a}{R}$ and the intercept is $\ln A$.Given:Slope $=-10^{3} K$Intercept $=2.303$$R=8314 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$The slope is related to the activation energy $E_a$ by:$-\;\frac{E_a}{R}=-10^{3}$Solving for $E_a$, we have:$E_a=10^{3} \cdot R$Substituting the value of $R$ :$E_a=10^{3} \cdot 8314 \mathrm{J} \mathrm{mol}^{-1}$To convert to kilojoules per mole, divide by 1000 :$E_a=\;\frac{8314 \times 10^{3}}{1000} \mathrm{kJ} \mathrm{mol}^{-1}=8314 \mathrm{kJ} \mathrm{mol}^{-1}$Therefore, the correct answer is:Option A: 8.314