Diborane and Chlorine react according to the reaction: B2H6(g) + 6Cl2(g) --> 2BC

Question

Diborane and Chlorine react according to the reaction:

B2H6(g) + 6Cl2(g) --> 2BCl3(g) + 6HCl(g) Delta H rxn = -1396kJ/mol

Suppose 1.00 mole of B2H6 is mixed with 3.00 moles of Cl2 at constant pressure of 1 atm and the reaction goes to completion.

a. Calculate the heat of the reaction at constant pressure, qp.

b. If the reaction is carried out at constant temperature of 298K and constant pressure of 1atm, calculate the work of the system in joules in this case. (1.00 mole B2H6 is mixed with 3.00 moles of Cl2) Assume ideal gases.

Explanation / Answer

for a constant pressure process, deltaH= Q

Diborane and Chlorine react according to the reaction:

B2H6(g) + 6Cl2(g) --> 2BCl3(g) + 6HCl(g) Delta H rxn = -1396kJ/mol

the reaction suggests 1 mole of B2H6 (g) reacts with 6 moles of Cl2 (g) to give rise to 2 mole of BCl3(g) and 6 moles of HCl(g) to give rise to 1396 KJ

molar ratio of B2H6: Cl2= 1:6 ( Theoretical), actual ratio given = 1:3

hence Cl2 will be limiting reactant. 6 moles of HCl when react gives rise to 1396 KJ

when 3 moles of Cl2 only react, q= deltHP= 1396/3=465.3 KJ

b) all the HCl gets consumed and BCl3 remains. total moles of reactants=3+0.5( reacted BCl3)+0.5( unreacted BCl3)= 3.5

moles of products= 1(BCl3)+3(HCl+0.5( unreacted BCl3)= 4.5

change in moles= 4.5-3.5= 1

hence work done = change in no of moles*R*T= 1*8.314*298 J ( R= gas constant = 8.314 J/mole.K)=2478 joules

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