1.) At what temperature will the rate constant for the reaction H2 + I2 to 2HI have the value 5.2×10−3 M^{-1}s^{-1}? (Assume k=5.4 *10^{-4}M}{-1)s}^{-1} at 599 K and k=2.8 * 10^{-2}M}^{-1}s}^{-1} at 683 K.)

2.)For the first-order reaction
N2O5 to 2NO2 + 1/2O2
t_{1/2}=22.5 h at 20 C and 1.5 h at 40 C.
What's the activation energy

1.) To find the temperature at which the rate constant for the reaction H2 + I2 to 2HI has a value of 5.2×10^(-3) M^(-1)s^(-1), we can use the Arrhenius equation:

k = A * e^(-Ea/RT)

where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin

We are given the values of k at two different temperatures: k = 5.4 *10^(-4) M^(-1)s^(-1) at 599 K and k = 2.8 * 10^(-2) M^(-1)s^(-1) at 683 K. We can use these values to set up two equations:

5.4 *10^(-4) = A * e^(-Ea/(8.314 * 599))
2.8 * 10^(-2) = A * e^(-Ea/(8.314 * 683))

Divide the second equation by the first equation:

2.8 * 10^(-2) / 5.4 *10^(-4) = e^(-Ea/(8.314 * 683)) / e^(-Ea/(8.314 * 599))

Taking the natural logarithm of both sides to eliminate the exponential terms:

ln(2.8 * 10^(-2) / 5.4 *10^(-4)) = (-Ea/(8.314 * 683)) - (-Ea/(8.314 * 599))

Simplifying:

ln(2.8 * 10^(-2) / 5.4 * 10^(-4)) = (Ea/(8.314 * 599)) - (Ea/(8.314 * 683))

Now, solve for Ea:

Ea = (8.314 * 599 * 683) * ln(2.8 * 10^(-2) / 5.4 * 10^(-4))

Ea ≈ 179,295 J/mol

Therefore, the activation energy for the reaction H2 + I2 to 2HI is approximately 179,295 J/mol.

2.) To find the activation energy for the first-order reaction N2O5 to 2NO2 + 1/2O2, we can use the Arrhenius equation:

k = A * e^(-Ea/RT)

where:
k = rate constant
A = pre-exponential factor
Ea = activation energy
R = gas constant (8.314 J/mol·K)
T = temperature in Kelvin

We are given the half-life of the reaction at two different temperatures: t_1/2 = 22.5 h at 20 °C (293 K) and t_1/2 = 1.5 h at 40 °C (313 K). The half-life can be related to the rate constant as follows:

t_1/2 = (ln(2))/(k)

We can rearrange the equation to solve for the rate constant:

k = (ln(2))/(t_1/2)

Now, substitute the given values to find the rate constant at each temperature:

k_1 = (ln(2))/(22.5) ≈ 0.0308 h^(-1)
k_2 = (ln(2))/(1.5) ≈ 0.4621 h^(-1)

Next, use the Arrhenius equation to set up an equation with two unknowns (A and Ea):

k_1 = A * e^(-Ea/(8.314 * 293))
k_2 = A * e^(-Ea/(8.314 * 313))

Divide the second equation by the first equation:

k_2/k_1 = e^(-Ea/(8.314 * 313)) / e^(-Ea/(8.314 * 293))

Taking the natural logarithm of both sides to eliminate the exponential terms:

ln(k_2/k_1) = (-Ea/(8.314 * 313)) - (-Ea/(8.314 * 293))

Simplifying:

ln(k_2/k_1) = (Ea/(8.314 * 293)) - (Ea/(8.314 * 313))

Now, solve for Ea:

Ea = (8.314 * 293 * 313) * ln(k_2/k_1)

Ea ≈ 44,234 J/mol

Therefore, the activation energy for the reaction N2O5 to 2NO2 + 1/2O2 is approximately 44,234 J/mol.

1.) To determine the temperature at which the rate constant for the reaction H2 + I2 → 2HI has a specific value, we can use the Arrhenius equation:

k = Ae^(-Ea/RT)

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/(mol K)),
- T is the temperature in Kelvin.

We are given the rate constants at two different temperatures: k = 5.4 * 10^(-4) M^(-1)s^(-1) at 599 K and k = 2.8 * 10^(-2) M^(-1)s^(-1) at 683 K.

To find the activation energy, we can rearrange the Arrhenius equation and solve for Ea:

ln(k1/k2) = (Ea/R) * (1/T2 - 1/T1)

Where:
- k1 and k2 are the rate constants at temperatures T1 and T2, respectively.

Substituting the given values, we have:

ln(5.4 * 10^(-4)/ 2.8 * 10^(-2)) = (Ea/8.314) * (1/683 - 1/599)

Simplifying and solving for Ea:

Ea = -8.314 * ln(5.4 * 10^(-4)/ 2.8 * 10^(-2)) / (1/683 - 1/599)

2.) To determine the activation energy for the first-order reaction N2O5 → 2NO2 + 1/2O2, we can use the Arrhenius equation as well:

k = Ae^(-Ea/RT)

Here, t_{1/2} represents the half-life of the reaction.

For a first-order reaction, the half-life (t_{1/2}) can be related to the rate constant (k) using the following equation:

t_{1/2} = (ln2) / k

We are given two different half-lives at temperatures 20°C and 40°C, which need to be converted to Kelvin.

To find the activation energy, we can rearrange the equation and solve for Ea:

ln(t_{1/2,1}/t_{1/2,2}) = (Ea/R) * (1/T2 - 1/T1)

Where t_{1/2,1} and t_{1/2,2} are the half-lives at temperatures T1 and T2, respectively.

Substituting the given values, we have:

ln(22.5/1.5) = (Ea/8.314) * (1/(273.15 + 20) - 1/(273.15 + 40))

Simplifying and solving for Ea:

Ea = -8.314 * ln(22.5/1.5) / (1/(273.15 + 40) - 1/(273.15 + 20))