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# Tutor profile: Reiner G.

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Reiner G.
Tutor for 8 years and English Teacher for a year
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## Questions

### Subject:Chemical Engineering

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Question:

The desintegration of propane is one road for the production of olefins. Suppose in a steady flow reactor two desintegration reactions take place. In one reaction, propane desintegrates into propylene and hydrogen; and in the other reaction it desintegrates into ethylene and methane. If the reactor is fed with 2 moles of propane, 1 mole propylene, and 10 moles of hydrogen, calculate the composition in equilibrium at 900K and 5 atm.

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Reiner G.

Note: All of the the properties used in this exercise are taken from the tables in "Introduction to Thermodynamics Van Ness" book 1. We need to write down and balance the reactions that take place $$C_3H_8(g)---> C_3H_6(g) + H_2(g)$$ (1) $$C_3H_8(g)---> C_2H_4(g) + CH_4(g)$$ (2) 2. Calculate the standard enthalpy of reaction, and standard entropy for each reaction. To do this we use the following. $$H_{reaction}=H_{prod}-H_{react}$$ --$$H_{prod}$$ and $$H_{react}$$ are the H of reaction of each component-- -- Remember to multiply the H of each component by it's stoichiometric coefficient-- -- To calculate entropy of formation for each component we use the equation $$G=H-TS$$ and we know standard means 25 C or 298.15K. Solving for S gives: $$S=G-H/(-T)$$ $$H_{(C_3H_8)}: -104680 J/mol$$ $$G_{(C_3H_8)}: -24290 J/mol$$ $$S_{(C_3H_8)}: -269.76 J/mol*K$$ $$H_{(C_3H_6)}: 19710 J/mol$$ $$G_{(C_3H_6)}: 62205 J/mol$$ $$S_{(C_3H_6)}: -142.60 J/mol*K$$ $$H_{(H_2)}: 0 J/mol$$ $$G_{(H_2)}: 0 J/mol$$ $$H_{(C_2H_4)}: 52510 J/mol$$ $$G_{(C_2H_4)}: 68460 J/mol$$ $$S_{(C_2H_4)}: -53.52 J/mol*K$$ $$H_{(CH_4)}: -74520 J/mol$$ $$G_{(CH_4)}: -50460 J/mol$$ $$S_{(CH_4)}: -80.74 J/mol$$ Now we calculate $$H_1^{298}, H_2^{298}, S_1^{298}, and S_2^{298}$$ $$H_1^{298}= [H_{(C_3H_6)} + H_{(H_2)}]-[H_{(C_3H_8)}]=(19710+0)-(-104680)=124390 J/mol$$ $$H_2^{298}=[H_{(C_2H_4)} + H_{(CH_4)}]-[H_{(C_3H_8)}]=[(52510+(-74520)]-(-104680)= 82670 J/mol$$ $$S_1^{298}= [S_{(C_3H_6)} + S_{(H_2)}]-[S_{(C_3H_8)}]=(-142.60-0)-(-269.76)=127.16 J/mol$$ $$S_2^{298}=[S_{(C_2H_4)} + S_{(CH_4)}]-[S_{(C_3H_8)}]=(-53.52-80.74))-(-269.76)= 135.5$$ 3. Calculate the equilibrium constant K at 900K a) First we obtain the Cp of reaction for each reaction. $$Cp_{reaction}=Cp_{prod}-Cp_{react}$$ $$Cp_{(C_3H_8)}: 10.08+0.2393T-(73.36*10^{-6})T^2$$ $$Cp_{(C_3H_6)}: 13.61+0.1888T-(57.49*10^{-6})T^2$$ $$Cp_{(H_2)}: 27.01+0.00351T+(0.083*10^5)/T^2$$ $$Cp_{(C_2H_4)}: 11.83+0.1197T-(36.51*10^{-6})T^2$$ $$Cp_{(CH_4)}: 14.19+0.0755T-(17.99*10^{-6})T^2$$ $$Cp_1: Cp_{(C_3H_6)}+Cp_{(H_2)}-Cp_{(C_3H_8)}$$ $$Cp_1=30.54-0.0470T+(15.87*10^{-6})T^2+(0.083*10^5)/T^2$$ $$Cp_2:Cp_{(C_2H_4)}+Cp_{(CH_4)}-Cp_{(CH_4)}$$ $$Cp_2=15.94-0.0441TT+(18.86*10^{-6})T^2$$ b)Now we use the the Cp of reaction to obtain Gibbs free energy at 900k We use the following equation: $$G^{900}=H^{298}+\int_{298}^{900} Cp_{reaction} dt - T[S^{298}+\int_{298}^{900} Cp_{reaction}/T dt$$ Which yields $$G_1^{900}=5014.44 J/mol$$ $$G_2^{900}=-39253.94 J/mol$$ We notice that G_1^{900} is a positive number, meaning that the reaction is spontaneous on the opposite direction. Since we have initial moles that can permit this opposite reaction to occur, (1) will be inverted and $$G_3^{900}=-G_1^{900}$$ This gives $$C_3H_6(g) + H_2(g)--->C_3H_8(g)$$ (3 $$G_3^{900}=-5014.44 J/mol$$ The equation to find the constant of equilibrium with Gibbs free energy is $$K=e^{[\frac{-G}{RT}]}$$ $$K_1^{900}=e^{[\frac{5014.44}{8.314*900}]}$$ $$K_1^{900}=1.95$$ $$K_2^{900}=e^{[\frac{39253.94}{8.314*900}]}$$ $$K_2^{900}=189.81$$ 5. We calculate the the breakthrough of reaction (3). a) First we write down the initial moles we have of each component $$n_{i_{C_3H_8}}=2 moles$$ $$n_{i_{C_2H_6}}=1 mol$$ $$n_{i_{H_2}}=10 moles$$ $$n_{i_{C_2H_4}}=0 moles$$ $$n_{i_{CH_4}}=0 moles$$ b) Now we write down the reaction breakthrough for each component in terms of y for breakthrough of reaction 3. A negative sign indicates consumption and a positive sign indicates production $$change_{C_2H_6}= -y mol$$ $$change_{H_2}= -y moles$$ $$change_{C_3H_8}= +ymoles$$ $$change_{C_2H_4}= 0 moles$$ $$change_{CH_4}= 0 moles$$ c) We determine the final moles of each component in equilibrium by adding up the initial moles and the change in moles. $$n_{f_{C_3H_8}}=2+y moles$$ $$n_{f_{C_2H_6}}= 1-y mol$$ $$n_{f_{H_2}}= 10-y moles$$ $$n_{f_{C_2H_4}}= 0 moles$$ $$n_{f_{CH_4}}= 0 moles$$ d) We add up all the final moles to determine the total amount of moles in equilibrium. $$n_{tot}=n_{f_{C_3H_8}}+n_{f_{C_2H_6}}+n_{f_{H_2}}+n_{f_{C_2H_4}}+n_{f_{CH_4}}$$ $$n_{tot}=2+y+1-y+10-y+0+0=13-y$$ e) The composition of each component will be the final moles of the component divided by the total moles. $$x_{C_3H_8}=[\frac{2+y}{13-y}]$$ $$x_{C_2H_6}= [\frac{1-y}{13-y}]$$ $$x_{H_2}= [\frac{10-y}{13-y}]$$ The equation that relates K of equilibrium with the composition is: $$K=K_x*K_p*K_f/K_{ref}$$ where $$K_p=[\frac{P_{prod}}{P_{react}}]$$ $$K_f$$ is one because we are working at low pressure $$K_{ref}$$ used is 1 $$K_x=[\frac{x_{prod}}{x_{react}}]$$ This yields $$Kp_1=[\frac{1}{P}]$$ $$K_1^{900}=[\frac{x_{C_3H_8}}{x_{C_3H_6}*x_{H_2}*P}]$$ $$1.95=[\frac{[\frac{2+y}{13-y}]}{[\frac{1-y}{13-y}]*[\frac{10-y}{13-y}]*5}]$$ $$y=0.6421$$ 6. We calculate the amount of final moles of each component that participated in reaction (3). $$n_{f_{C_3H_8}}=2+y=2+0.6421=2.6421$$moles $$n_{f_{C_2H_6}}= 1-y=1-0.6421=0.3579$$ moles $$n_{f_{H_2}}= 10-y=10-0.6421=9.3579$$ moles 7. We calculate the breakthrough of consecutive reaction 2. a) First we write down the initial moles we have of each component $$n_{i_{C_3H_8}}=2.6421 moles$$ $$n_{i_{C_2H_6}}=0.3579mol$$ $$n_{i_{H_2}}=9.3579 moles$$ $$n_{i_{C_2H_4}}=0 moles$$ $$n_{i_{CH_4}}=0 moles$$ b) Now we write down the reaction breakthrough for each component in terms of z for breakthrough of reaction 2. A negative sign indicates consumption and a positive sign indicates production $$change_{C_2H_6}= -z mol$$ $$change_{H_2}= 0 moles$$ $$change_{C_3H_8}= 0 moles$$ $$change_{C_2H_4}= +z moles$$ $$change_{CH_4}= +z moles$$ c) We determine the final moles of each component in equilibrium by adding up the initial moles and the change in moles. $$n_{f_{C_3H_8}}=2.6421-z moles$$ $$n_{f_{C_2H_6}}= 0.3579 mol$$ $$n_{f_{H_2}}= 9.3579 moles$$ $$n_{f_{C_2H_4}}= z moles$$ $$n_{f_{CH_4}}= z moles$$ d) We add up all the final moles to determine the total amount of moles in equilibrium. $$n_{tot}=n_{f_{C_3H_8}}+n_{f_{C_2H_6}}+n_{f_{H_2}}+n_{f_{C_2H_4}}+n_{f_{CH_4}}$$ $$n_{tot}=2.6421-z+0.3579+9.3579+z+z=12.3579+z$$ e) The composition of each component will be the final moles of the component divided by the total moles. $$x_{C_3H_8}=[\frac{2.6421-z}{12.3579+z}]$$ $$x_{C_2H_6}= [\frac{0.3579}{12.3579+z}]$$ $$x_{H_2}= [\frac{9.3579}{12.3579+z}]$$ $$x_{C_2H_4}= [\frac{z}{12.3579+z}]$$ $$x_{CH_4}=[\frac{z}{12.3579+z}]$$ Since $$C_2H_6$$ and $$H_2$$ do not participate in the reaction, they are not considered for the calculation of the reaction breakthrough; but are considered in the total mole in equilibrium and for the composition calculation. $$Kp_2=P$$ $$K_2^{900}=[\frac{[\frac{z}{12.3579+z}]*[\frac{z}{12.3579+z}]*5}{[\frac{2.642-z}{12.3579+z}]}]$$ $$z=2.6416$$ 8. Calculate the composition of each component in equilibrium. $$x_{C_3H_8}=3.33*10^{-5}$$ $$x_{C_2H_6}= 0.0239$$ $$x_{H_2}=0.6239$$ $$x_{C_2H_4}=0.1761]$$ $$x_{CH_4}=0.1761$$

### Subject:Algebra

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Question:

Find the equation to the line that is perpendicular to the line 3x+ 2y = 12 and passes through the points (-3,-6).

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Reiner G.

1. We pass the equation of the line to the slope-intercept form (y=mx+b) 2y=-3x+12 $$y=(-3/2)x+6$$ 2. When lines are perpendicular to each other, the product of their slopes is -1 (m1*m2=-1); so we find the m2 by finding the reciprocate of m1 and changing the sign. $$m1= (-3/2)$$ $$m2= 2/3$$ 3. We use the slope and the points given to find the equation of the line. $$y-y1=m2(x-x1)$$ $$y-(-6)=(2/3)(x-(-3))$$ $$y+6=(2/3)(x+3)$$ We use the distributive property in (2/3)(x+3) $$y=(2/3)x+(2/3)(3)-6$$ $$y=(2/3)x+(2)-6$$ $$y=(2/3)x-4$$

### Subject:Physics

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Question:

There is a system of three blocks linked by frictionless strings and pulleys. Block 1 has a mass m1= 10kg, block 2 has a mass m2=30kg, and block 3 has a mass m3=50kg. Block 1 is hanging on the left side of a table on a string attached to block 2, which is on the table. Block 3 is hanging on the right side of the table on a string also attached to block 3. a) Find the magnitude of the acceleration of all three blocks. b) Find the magnitude of the tension of the string between block 1 and 2. c) Find the magnitude of the tension of the string between block 2 and 3.

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Reiner G.

1. Draw the free body diagram including all the forces that act on each of the blocks. 2. We apply Newton's second law. - Since m3 has the highest mass, we assume m1 is accelerating up, m2 is accelerating to the right, and m3 is accelerating down. - Equation for m1 $$T1 - w1= m1*a$$ - Equation for m2 $$T2- T'1 + N - w2= m2*a$$ T1=T'1, and N=w2, so equation for m2 becomes -----> $$T2-T1=m2*a$$ - Equation for m3 $$T'2 - w3= -m3*a$$ T'2=T2, so equation for m3 becomes ----> $$T2-w3=-m3*a$$ 3. We combine all equations by solving equation 1 for T1, solving equation 3 for T2, and substituting both in equation 2. - $$T1=m1*a + w1$$ -$$T2=-m3*a + w3$$ substitung both in equation 2 yields $$-m3*a + w3 - (m1*a + w1)= m2*a$$ we know that $$wi= mi * g$$ our equation thus becomes $$-m3*a + m3*g - m1*a - m1*g=m2*a$$ now we solve for a $$a=g(m3-m1)/(m2+m1+m3)$$ plugging in our magnitudes m1=10 kg m2=30 kg m3=50 kg g= 9.81 m/s^2 a) $$a= 9.81*(50-10)/(10+30+50)= 4.36 m/s^2$$ b) $$T1= 10*4.36+10*9.81=141.7 N$$ c)$$T2=50*4.36+50*9.81=708.5 N$$

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