Question

which of the following reactions k310/k300 would be maximum

Answer 1

The complete question is,

which of the following reactions k310/k300 would be maximum?

a. A + B → C, Ea = 500 kJ  

b. X + Y → Z, Ea = 40 kJ

c. P + Q → R, Ea = 600 kJ

d. E + F → G, Ea = 100 kJ

Given:

a. A + B → C, Ea = 500 kJ  

b. X + Y → Z, Ea = 40 kJ

c. P + Q → R, Ea = 600 kJ

d. E + F → G, Ea = 100kJ

To find:

Which reactions k310/k300 would be maximum?

Solution:

We use the formula,

k = Ae^{-Ea/RT}

For T = 300 K

k_{300} = Ae^{-Ea/R(300)}

For T = 310 K

k_{310} = Ae^{-Ea/R(310)}

Now consider,

k_{310} /k_{300} = Ae^{-Ea/R(310)} / Ae^{-Ea/R(300)}

= e^{-Ea/R(310)} / e^{-Ea/R(300)}

= e^{-Ea/R (1/310 - 1/300)}

= e^{-Ea/R (300 - 310) / (300 × 310)}

= e^{-Ea/R (-10) / (300 × 310)}

= e^{Ea/R (10) / (300 × 310)}

= e^{Ea / (R × 300 × 31)}

k_{310} /k_{300} = Ea maximum

Option d. E + F → G, Ea = 100 kJ is correct.

Answer 2

d. E + F → G, Ea = 100 kJ

Explanation:

The question is incomplete. Please find attached the choice of answers for this question. The next picture shows how the solution is derived. It would be easy to understand in the picture than writing it in text here because of formatting. E + F → G, Ea = 100 kJ has the highest energy.

Option D is the answer.