Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae ------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
Answer: The exit temperature of the gas in deg C is .
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is .