ratedetrmining-step: 3X(g) + 2Y(g) = 4Z(g)
When 2 moles of gas X and 2 moles of gas Y are placed in a 5.0L vessel, the initial rate of the rxn is found to be 0.45 (mol/L x min. What is the rate constant?
P.S: I think k=1.875(L/mol x s) but im not sure.
To determine the rate constant (k) for a reaction, you need to use the rate law expression and the initial concentrations of the reactants involved.
In your case, the balanced equation for the reaction is:
3X(g) + 2Y(g) = 4Z(g)
The rate law expression for this reaction is:
Rate = k[X]^a[Y]^b
where [X] and [Y] are the concentrations of the reactants X and Y, respectively, and 'a' and 'b' are the stoichiometric coefficients for X and Y in the balanced equation.
From the balanced equation, we can see that 'a' is 3 and 'b' is 2, so the rate law expression becomes:
Rate = k[X]^3[Y]^2
Given that the initial concentrations of X and Y are both 2 moles/L, you can substitute these values and the initial rate into the rate law expression.
0.45 (mol/L x min) = k(2 mol/L)^3(2 mol/L)^2
0.45 = k(8 mol^3/L^3)(4 mol^2/L^2)
0.45 = k(256 mol^5/L^5)
Now, solve for the rate constant (k):
k = 0.45 / (256 mol^5/L^5)
Evaluate the expression:
k ≈ 1.7578 x 10^-7 (mol^4/L^4 x min)
So, the rate constant (k) for this reaction is approximately 1.7578 x 10^-7 (mol^4/L^4 x min).