0 votes
30 views
in ElectroChemistry by (138k points)
What is the activation energy for the reaction between 2 M of HCl and 2 M of NaOH at a temperature of 25°C when a voltage of 1.5 V is applied?

1 Answer

0 votes
by (138k points)
To calculate the activation energy for the reaction between 2 M HCl and 2 M NaOH at 25C with a voltage of 1.5 V applied, we can use the Arrhenius equation:k = A * exp -Ea /  R * T  where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant  8.314 J/molK , and T is the temperature in Kelvin  25C = 298 K .However, we do not have enough information to directly calculate the activation energy  Ea  using the Arrhenius equation. We would need the rate constant  k  or the pre-exponential factor  A  for the reaction.Alternatively, if we assume that the applied voltage is used to overcome the activation energy barrier, we can use the following equation:Ea = n * F * Vwhere n is the number of electrons transferred in the reaction, F is the Faraday constant  96,485 C/mol , and V is the applied voltage  1.5 V .For the reaction between HCl and NaOH, the balanced equation is:HCl + NaOH  NaCl + H2OIn this reaction, one electron is transferred  n = 1 . Therefore, we can calculate the activation energy as follows:Ea = 1 * 96,485 C/mol * 1.5 VEa = 144,727.5 J/molSo, the activation energy for the reaction between 2 M HCl and 2 M NaOH at 25C with a voltage of 1.5 V applied is approximately 144,727.5 J/mol. However, this calculation assumes that the applied voltage is solely responsible for overcoming the activation energy barrier, which may not be entirely accurate in a real-world scenario.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...