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Problem: The activation energy (Ea) for a certain chemical reaction is 50.0 kJ/mol. When the temperature is raised from 300 K to 310 K, what happens to the rate constant? Given R = 8.314 J/(mol K). Step 1: Identify the given values from the problem statement: Ea = 50.0 kJ/mol = 50,000 J/mol, T1 = 300 K, T2 = 310 K, and R = 8.314 J/(mol K). Step 2: Write down the two-point Arrhenius equation for calculating the change in rate constants at two different temperatures: ln(k2 / k1) = (Ea / R) * ((T2 - T1) / (T1 * T2)). Step 3: Substitute the known values into the right side of the equation: ln(k2 / k1) = (50,000 / 8.314) * ((310 - 300) / (300 * 310)). Step 4: Simplify the numbers inside the parentheses and fraction: ln(k2 / k1) = 6013.95 * (10 / 93000) = 6013.95 * 0.0001075 = 0.6465. Step 5: Take the inverse natural logarithm (exponential function) of both sides to find the ratio of the rate constants: k2 / k1 = e^(0.6465) = 1.91. Conclusion: The rate constant increases by about 1.91 times when the temperature increases by just 10 Kelvin.