A process utilizes the reaction of 2-chloro-2-methylpropane (C4HgCl) with an aqu

Question

A process utilizes the reaction of 2-chloro-2-methylpropane (C4HgCl) with an aqueous solution of sodium hydroxide (NaOH) to produce 2-methlypropene (C4HB), along with sodium chloride (NaCl) and water (H20). Unfortunately, the initial reactants can engage in a side reaction that produces 2-methyl-2-propanol (C4H00) and sodium chloride. Given the flow rates in the flow diagram and by following each step below, determine Q, the flow rate of heat entering the reactor that keeps the outlet temperature at 25 C. Ignore any contribution caused by heats of mixing. Also see below for a table of heats of formation of each species. Compound r (kJ/mol) -191.10 T- 25°C T 25°C Reactor s 10.3 mol c4HB/hr NaOH -469.40 0 0 n1 16.1 mol C4H9CIhr 285.84 16.1 mol NaoH/hr mol C4HOOIhr H20 ns inb mol NaCUhr 805 mol H20/hr -16.90 C4H8 m mole H20/hr Q kJ/hr -74.70 C4H10O NaCl -407.27 What is ns, the flow rate of C4H10O out of the reactor? Scroll down for more parts. Number mol C 10 e Previous 8 Give Up & View Solution 0 Check Answer e Next Exit

Explanation / Answer

The reactions are

C4H9Cl +NaOH---à C4H8+NaCl +H2O    (1)

C4H9Cl +NaOH---àC4H9OH+ NaCl          (2)

1 mole of C4H9Cl gives 1 mole of C4H8. But noles of C4H8 formed = 10.3 mol/hr

C4H8 formed = 10.3 mol/hr, moles of C4H9Cl used = 16.1 mol/hr

Moles of C4H9Cl used for formation of C4H9OH= 16.1-10.3=5.8 moles/hr

Moles of C4H9OH formed,n5 =5.8 moles/hr , moles of NaCl formed ,n6= moles of NaCl formed from reaction -1 + moles of NaCl formed from reaction -2 =   10.3+5.8 =16.1 moles/hr

Moles of water , n7= moles of water at inlet + moles of water formed from reaction -1 = 805+10.30=815.3 moles/hr

Heat   input   = heat out + Q ( heat removed from reactor) , Q= heat of reactants- heat out from products

=16.1*(-191.1)+ 16.1*(-469.40)+ 805*(-285.84) – { 10.3*(-16.90) + 16.1*(-407.27) + 5.8*(-74.70)+815.3*(-285.84)} =-240735+240210 =-525.51 Kj

Q= heat added to reactor = 525.51 KJ/hr

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