Final answer:
To find out how much faster a reaction is at 343 K than at 323 K, given the activation energy (Ea) is 42.2 kJ/mol, one would use the Arrhenius equation and compare the rate constants at both temperatures. The exact rate increase requires the frequency factor which is not provided, yet the reaction will undoubtedly be faster at the higher temperature due to the temperature's exponential effect on the rate constant.
Step-by-step explanation:
The student asked how much faster a reaction is at 343 K compared to 323 K with a known activation energy (Ea) of 42.2 kJ/mol. To calculate this, we can use the Arrhenius equation, k = Ae-Ea/RT, where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the ideal gas constant (8.314 J/mol/K), and T is the temperature in Kelvin.
First, we need to convert the activation energy from kJ/mol to J/mol by multiplying it by 1,000, so Ea = 42,200 J/mol. Then we apply the Arrhenius equation to calculate the rate constants (k) for both temperatures. Once we obtain the values of k at 343 K and 323 K, we can divide the former by the latter to find out how much faster the reaction is at the higher temperature.
We can't provide an exact numerical answer without knowing the frequency factor (A), but the process involves comparing the exponential terms for both temperatures to determine the relative increase in reaction rate. As the temperature increases, the exponential term becomes less negative, resulting in a larger value for k and a faster reaction rate.