At 1650 °C the value of Kc for the equilibrium represented by the balanced equation is 2.38 x 10-1. Calculate the value of Kp at the same temperature. Express answer in scientific notation.
CO(g)+H2O(g) = CO2(g)+H2(g)
___ x 10 ____
As I remember,
Kp=Kc * (RT)n where n is equal to the sum of moles on products side minus total moles on reactants side.
I would make a quick guess it is 0.238. Why?
Kp = Kc(RT)^delta n
where dn = nproducts - nreactants.
I see 2 mols on the right and 2 mols on the left which means dn = 0, RT raised to the zero power is 1; bingo.
To calculate the value of Kp at the same temperature, we need to use the relationship between Kc and Kp for a gaseous equilibrium.
The equation is given as:
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
The expression for Kp is given as:
Kp = Kc(RT)^(∆n)
Where Kp is the equilibrium constant in terms of partial pressure, Kc is the equilibrium constant in terms of concentration, R is the gas constant (0.08206 L⋅atm/(mol⋅K)), T is the temperature in Kelvin, and ∆n is the change in moles during the reaction (sum of the moles of products minus the sum of the moles of reactants).
In this case, we need to determine the value of Kp at 1650 °C. First, we need to convert the temperature to Kelvin by adding 273.15 to it:
1650 °C + 273.15 = 1923.15 K
Next, we determine the change in moles (∆n) for the reaction:
∆n = (moles of products) - (moles of reactants)
∆n = (1 + 1) - (1 + 1) = 0
Since ∆n is 0, the equation for Kp simplifies to:
Kp = Kc(RT)^0
Kp = Kc
Now, substitute the given value of Kc into the equation:
Kp = 2.38 x 10^(-1)
Therefore, the value of Kp at the same temperature is 2.38 x 10^(-1), expressed in scientific notation.