Substance P is synthesized from substances X and Y in a constant-flow reactor which has two feeds fo — Physical Chemistry Chemistry Question

Since this is a reactor with ideal stirring, the concentrations of substances in the output flow are equal to the concentrations inside the reactor. In a stationary state, the concentrations and quantities of substances in the reactor are constant. Consider the material balance with respect to X, Y and P.
Stationary conditions are:
∆νX,R / ∆t = 0; ∆νY,R / ∆t = 0; ∆νP,R / ∆t = 0, (1)
where ∆νX,R, ∆νY,R, ∆νP,R are the changes of the quantities for the substances X, Y and P in the reactor during time ∆t. The quantity of the substance in the reactor may change due to input flow, chemical reaction, and output flow:
(∆νX,R / ∆t) = (∆νX,R / ∆t)input + (∆νX,R / ∆t)reaction + (∆νX,R / ∆t)output (2)
The same is true for Y and P.
Input flow rates of the substances are
(∆νX,R / ∆t)input = fX cX,I ; (∆νY,R / ∆t)input = fY cY,I ; (∆νP,R / ∆t)input = 0, (3)
where fX and fY are the input volumetric flows of the solutions of X and Y, cX,I and cY,I – concentrations of X and Y in the respective solutions.
Let the balanced reaction equation be
nX X + nY Y = nP P
where nX, nY and nP are the stoichiometric coefficients for the corresponding substances.
Due to a chemical reaction the quantities of the substances in the reactor change with the rates
(∆νX,R / ∆t)reaction = - nX r VR; (∆νY,R / ∆t)reaction = - nY r VR; (∆νP,R / ∆t)reaction = nP r VR, (4)
where r – the reaction rate, VR – the reactor volume.
The output flows of the substances are:
(∆νX,R / ∆t)output = - fO cX,R ; (∆νY,R / ∆t)output = - fO cY,R ; (∆νP,R / ∆t)output = - fO cP,R , (5)
where fO is the volumetric output flow, cX,R, cY,R and cP,R – the concentrations of substances X, Y and P in the reactor. Since the process is stationary and the reaction proceeds in the liquid phase, the output volumetric flow equals the sum of input volumetric flows:
fO = fX + fY (6)
Thus the material balance equations (2) considering expressions (1) and (3)^-(6) are
∆νX,R / ∆t = fX cX,I - nX r VR - cX,R (fX + fY) = 0
∆νY,R / ∆t = fY cY,I - nY r VR - cY,R (fX + fY) = 0
∆νP,R / ∆t = nP r VR - cP,R (fX + fY) = 0
Hence
nX r VR = fX cX,I - cX,R (fX + fY)
nY r VR = fY cY,I - cY,R (fX + fY)
nP r VR = cP,R (fX + fY)

Exp. no. | nXrVR, mol/s | nYrVR, mol/s | nPrVR, mol/s | nX:nY:nP
1 | 10.02 | 20.04 | 10.02 | 1:2:1
2 | 10.04 | 20.07 | 10.05 | 1:2:1
3 | 15.73 | 31.47 | 15.72 | 1:2:1
4 | 19.68 | 39.34 | 19.68 | 1:2:1

Hence the balanced reaction equation is
X + 2 Y = P

Now consider the rate dependence on concentrations
Exp. no. | cX,R, mol/m₃ | cY,R, mol/m₃ | cP,R, mol/m₃ | r VR, mol/s
1 | 299 | 48.2 | 501 | 10.02
2 | 732 | 30.9 | 335 | 10.04
3 | 8.87 | 351 | 524 | 15.73
4 | 308 | 66.6 | 492 | 19.68

The rate law is r = k cX,R^x cY,R^y cP,R^p
or, after multiplying by reactor volume, r VR = k VR cX,R^x cY,R^y cP,R^p
Take the logarithm of both parts of the equation
ln(rVR) = ln(kVR) + x ln(cX,R) + y ln(cY,R) + p ln(cP,R) (7)

The coefficients in this equation are given in the table below:
Exp. no. | ln cX,R | ln cY,R | ln cP,R | ln (rVR)
1 | 5.70 | 3.88 | 6.22 | 2.30
2 | 6.60 | 3.43 | 5.81 | 2.31
3 | 2.18 | 5.86 | 6.26 | 2.76
4 | 5.73 | 4.20 | 6.20 | 2.98

Solving the system of equations (7) we get:
x = 1.00; y = 2.00; p = 0.01; ln(kVR) = -11.20
Hence the orders of the reaction are one in X, two in Y, and zero in P. The product kVR is:
k VR = exp(-11.20) = 1.37 · 10^-5 m9mol^-2s^-1.

One of the possible mechanisms that match the obtained rate law is:
X + Y ↔ I (fast)
I + Y → P (slow, rate-determining)

Summarizing, the obtained results are:
♦ the reaction equation: X + 2 Y = P;
♦ the reaction orders: 1, 2, and 0 with respect to X, Y, and P respectively;
♦ the product of the rate constant and reactor volume: k VR = 1.37 · 10^-5 m9⋅mol^-2⋅s^-1.