For a reaction with an activation energy of 50 kJ/mol, what is the effect of a 10 kJ/mol increase in activation energy on the rate constant at a constant temperature?
Practice Questions
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Q1
For a reaction with an activation energy of 50 kJ/mol, what is the effect of a 10 kJ/mol increase in activation energy on the rate constant at a constant temperature?
Rate constant increases
Rate constant decreases
Rate constant remains the same
Rate constant becomes zero
An increase in activation energy decreases the rate constant according to the Arrhenius equation, k = Ae^(-Ea/RT), where an increase in Ea results in a smaller k.
Questions & Step-by-step Solutions
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Q
Q: For a reaction with an activation energy of 50 kJ/mol, what is the effect of a 10 kJ/mol increase in activation energy on the rate constant at a constant temperature?
Solution: An increase in activation energy decreases the rate constant according to the Arrhenius equation, k = Ae^(-Ea/RT), where an increase in Ea results in a smaller k.
Steps: 6
Step 1: Understand what activation energy (Ea) is. It is the minimum energy required for a reaction to occur.
Step 2: Know that the rate constant (k) is a number that tells us how fast a reaction happens.
Step 3: Familiarize yourself with the Arrhenius equation: k = A * e^(-Ea/RT). Here, A is a constant, e is the base of natural logarithms, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
Step 4: Recognize that if the activation energy (Ea) increases, the exponent (-Ea/RT) becomes more negative.
Step 5: Understand that a more negative exponent means that e^(-Ea/RT) becomes smaller.
Step 6: Conclude that if the value of k decreases, the reaction rate slows down when activation energy increases.
