A 5.0-g Sample of KBr at 25.0 degrees celsius dissolves in 25.0 degrees celsius 25.0 mL of water also at 25.0 degrees celsius. The final equilibrium temperature of the resulting soltution is 18.1 degrees celsius. What is the enthalpy of of solution in kilojoules per mole of KBr?
2 answers
Try reading your question. It needs to be cleaned up. It makes no sense as is.
-q = dH ( dH = change in enthalpy Kj/mol)
Use q=mc(Tf-Ti) and then convert to deltaH
q=mc(Tf-Ti)
q =?
c= 4.814 J/g*C use specific heat of water since its in aqueous sol and we are not told differently
m= total mass= mass KBr + mass water
mass KBr= 5.0 g
*mass of water from density =1g/mL so
(25.0ml) (1g/ml)=25.0g water
m = 25.0g water + 5.0 g KBr = 30.0g
Tf-Ti= 18.1 - 25.0 = -6.9C
q= (30g) (4.184 J/gC)(-6.9C)
q= -1443 J
-q = delta H kJ/mol KBr
dH= (-q Joules)( 1 kJ/1000 Joules) / mol KBr
*mol KBr = (5.0 g KBr)(1mol KBr) /119g KBr)= 0.042 mol KBr
dH = -(-1443 J) (1kJ/1000J)/ 0.042 mol KBr)
dH= 34.4 kJ/mol KBr
Use q=mc(Tf-Ti) and then convert to deltaH
q=mc(Tf-Ti)
q =?
c= 4.814 J/g*C use specific heat of water since its in aqueous sol and we are not told differently
m= total mass= mass KBr + mass water
mass KBr= 5.0 g
*mass of water from density =1g/mL so
(25.0ml) (1g/ml)=25.0g water
m = 25.0g water + 5.0 g KBr = 30.0g
Tf-Ti= 18.1 - 25.0 = -6.9C
q= (30g) (4.184 J/gC)(-6.9C)
q= -1443 J
-q = delta H kJ/mol KBr
dH= (-q Joules)( 1 kJ/1000 Joules) / mol KBr
*mol KBr = (5.0 g KBr)(1mol KBr) /119g KBr)= 0.042 mol KBr
dH = -(-1443 J) (1kJ/1000J)/ 0.042 mol KBr)
dH= 34.4 kJ/mol KBr