The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at $25^{\circ }\mathrm{C}$ are $3.0\times 10^{-4}{\mathrm{s}}^{-1}$, $104.4\mathrm{kJ}/\mathrm{mol}$ and $6.0\times 10^{14}{\mathrm{s}}^{-1}$, respec-tively. The value of the rate constant as $T\to \infty$, is(a) $2.0\times 10^{18}{\mathrm{s}}^{-1}$(b) $6.0\times 10^{14}{\mathrm{s}}^{-1}$(c) infinity(d) $3.0\times 10^{-4}{\mathrm{s}}^{-1}$

Step by Step Solution: Step 1. Given rate constant, $k=3.0\times 10^{-4}{s}^{-1}$, activation energy, $E_a=104.4kJ/mol$ and Arrhenius parameter, $A=6.0\times 10^{14}{s}^{-1}$ at $T=25^{\circ }C$ Step 2. To find the value of rate constant as $T\to \infty$ Step 3. We can use the Arrhenius equation which relates rate constant with temperature as follows: Step 4. $k=A{e}^{-\frac{E_a}{RT}}$ Step 5. Taking the natural log both sides of equation (4), we get: Step 6. $ln(k)=ln(A)-\frac{E_a}{RT}$ Step 7. As temperature approaches infinity, the value of $1/T$ approaches zero. We can therefore write the equation as follows: Step 8. $ln(k)=ln(A)-\frac{E_a}{R\times \infty }$ Step 9. $ln(k)=ln(A)-0$ Step 10. $ln(k)=ln(A)$ Step 11. Expanding both terms using the logarithmic rule of product, we get: Step 12. $ln(k)=ln(6.0\times 10^{14})$ Step 13. Solving for $k$, we get: Step 14. $k=e^{ln(6.0\times 10^{14})}\approx 6.0\times 10^{14}{s}^{-1}$ Final Answer: (b) 6.0 \times 10^{14}~s^{-1}