The activation energy for the isomerization of methyl isonitrile is 160 kJ/mol. Calculate the fraction of methyl isonitrile molecules that have an energy of 160 or greater at 500 K? 510 K? What is the ratio of the fraction at 510K to that at 500K?
I missed this lesson today and would really appreciate someone explaining how to solve this. Thanks =]
Use the equation for activation complex: fraction=e^-[activation energy/8.314J/mol*temperature
(also, simply put: f=e^-(Ea/Rt) )
(I'm a chem student learning/studying this at the moment)
To calculate the fraction of methyl isonitrile molecules that have an energy of 160 kJ/mol or greater at a specific temperature, we can use the Boltzmann distribution equation:
f(E) = e^((-Ea)/(RT))
where:
f(E) is the fraction of molecules with energy E or greater,
Ea is the activation energy,
R is the ideal gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin.
First, we need to convert the activation energy from kJ/mol to J/mol:
Ea = 160 kJ/mol = 160,000 J/mol
Now, let's calculate the fraction of molecules at 500 K:
f(500K) = e^((-160,000 J/mol) / ((8.314 J/(mol·K)) * 500 K))
Calculate the exponent:
(-160000 J/mol) / ((8.314 J/(mol·K)) * 500 K) ≈ -38.45
Now, calculate the fraction:
f(500K) = e^(-38.45) ≈ 1.930 × 10^(-17)
Similarly, calculate the fraction of molecules at 510 K:
f(510K) = e^((-160,000 J/mol) / ((8.314 J/(mol·K)) * 510 K))
Calculate the exponent:
(-160000 J/mol) / ((8.314 J/(mol·K)) * 510 K) ≈ -37.92
Now, calculate the fraction:
f(510K) = e^(-37.92) ≈ 8.556 × 10^(-17)
Finally, calculate the ratio of the fraction at 510 K to that at 500 K:
Ratio = f(510K) / f(500K) ≈ (8.556 × 10^(-17)) / (1.930 × 10^(-17)) ≈ 4.435
So, the ratio of the fraction at 510 K to that at 500 K is approximately 4.435.
To calculate the fraction of molecules that have an energy of 160 kJ/mol or greater, we need to use the Boltzmann distribution equation. The equation is given as:
f(E) = e^(-E/RT),
where f(E) represents the fraction of molecules with an energy E or greater, e is the base of the natural logarithm, E is the energy, R is the ideal gas constant (8.314 J/(mol·K)), and T is the temperature in Kelvin.
In this case, we need to convert the energy from kJ/mol to J/mol by multiplying it by 1000.
Now, let's calculate the fraction of molecules that have an energy of 160 kJ/mol or greater at 500 K:
E = 160 kJ/mol = 160,000 J/mol
T = 500 K
f(E) = e^(-E/RT) = e^(-160,000 J/mol / (8.314 J/(mol·K) * 500 K))
To calculate this, we can use a scientific calculator or an online calculator that supports exponential functions. Plugging in the values and calculating, we find:
f(E) at 500 K ≈ 2.47332 × 10^(-14)
Similarly, we can calculate the fraction at 510 K:
T = 510 K
f(E) at 510 K = e^(-160,000 J/mol / (8.314 J/(mol·K) * 510 K))
f(E) at 510 K ≈ 2.30577 × 10^(-14)
To find the ratio of the fraction at 510 K to that at 500 K:
Ratio = f(E) at 510 K / f(E) at 500 K
Ratio ≈ (2.30577 × 10^(-14)) / (2.47332 × 10^(-14))
Ratio ≈ 0.9323
Therefore, the ratio of the fraction at 510 K to that at 500 K is approximately 0.9323.