Calculate the value of Kp at 227oC for the equilibrium: 3 A(g) ⇌ B(g) + D(g)
Kc = 5.15
Kp = Kc(RT)^delta n where delta n = number of mols products - number of moles reactants
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To find the value of Kp at 227oC, we need to use the equation relating Kp and Kc. The equation is:
Kp = Kc(RT)^(∆n)
Where:
Kp is the equilibrium constant in terms of partial pressures,
Kc is the equilibrium constant in terms of concentrations,
R is the ideal gas constant (0.0821 L.atm/(mol.K)),
T is the temperature in Kelvin,
Δn is the change in moles of gaseous products minus the change in moles of gaseous reactants.
In this case, the balanced equation is:
3 A(g) ⇌ B(g) + D(g)
We see that there is no change in the number of moles of gaseous products or reactants. Therefore, Δn = 0.
We are given Kc = 5.15
And we need to determine Kp at 227oC.
To convert 227oC to Kelvin, we add 273:
T = 227 + 273 = 500 K
Using the equation Kp = Kc(RT)^(∆n), and substituting the given values, we get:
Kp = 5.15(0.0821)(500)^0
Since ∆n = 0, the value of (∆n)^0 is equal to 1.
Therefore, the value of Kp at 227oC is:
Kp = 5.15(0.0821)(1)
Kp = 0.421
To calculate the value of Kp at a given temperature, we need to use the equation that relates Kp and Kc:
Kp = Kc * (RT)^(Δn)
Where:
- Kp is the equilibrium constant in terms of partial pressures
- Kc is the equilibrium constant in terms of molar concentrations
- R is the ideal gas constant (0.0821 L.atm/mol.K)
- T is the temperature in Kelvin
- Δn is the difference in the number of moles of gaseous products and reactants (Δn = np - nr)
In this case, the equation is:
3 A(g) ⇌ B(g) + D(g)
Δn = np - nr = (1 + 1) - 3 = -1
Given that Kc = 5.15 and the temperature is 227°C, we need to convert it to Kelvin:
T(K) = 227°C + 273.15 = 500.15 K
Now, we can substitute the values into the equation to calculate Kp:
Kp = Kc * (RT)^(Δn) = 5.15 * (0.0821 L.atm/mol.K)^(1) = 0.4231 L^(1).atm^(1)/mol^(1)
So, the value of Kp at 227°C for the given equilibrium is 0.4231 L^(1).atm^(1)/mol^(1).