Help with enthalpy question! (1 Viewer)

[Munkiie]

I really need help answering these questions... I don't really know what to do first =.=
Can someone please talk me through the steps I should take to do these questions??
Thank you in advance

1. When 50ml of 1.0mol/L potassium hydroxide solution is added to 50ml of 1.0mol/L hydrochloric acid solution, 2.8kj of heat energy is given out. Calculate the enthalpy change for the reaction: H + OH --> h20

2. Given H + OH --> h20 ΔH= -57kj/mol
Calculate the heat released when 50ml 0.1mol/L Ba(oh)2 reacts with 50ml 0.1mol/L h2so4

 

[DamTameNaken]

1. 50 mL of 1.0mol/L solution is the equivalent of 0.05 mol

0.05 mol H reacts with 0.05 mol OH ---> 0.05 mol H2O + ΔH-2.8kJ

The reason it's a negative is energy is taken AWAY from the bonds to produce heat.

Because the above reaction is in terms of 0.05 mol and enthalpy change is measured in moles we can just multiply everything by 20

1 mol H + 1 mol OH ---> 1 mol H2O + ΔH-56kJ

we can get rid of all the '1 mol' stuff and just write

H + OH ---> H20 + ΔH-56kJ

the enthalpy change is therefore 56kJ



2. 50 mL of 0.1 mol/L = 0.005 mol

For this reaction ignore the barium and sulfate as we are only dealing with the water half equation.

0.005 mol (OH)2 + 0.005 mol H2 ---> 0.005 mol 2(H2O)

Above is the overall reaction notice that for the H2O we form 0.01 mol of H2O. In a similar way we can write the H2 as 2xH and (OH)2 as 2xOH

0.01 mol OH + 0.01 mol H2 ----> 0.01 mol H2O

if the equation was with 1 mol of each substance then the heat released would be the energy equivalant of 57kJ, but as we're dealing with 1% the concentration we divide it by 100, i.e. 0.57 kJ of heat is produced.

Now you can go fuirther and workout the temperature change of the solution.

ΔH = mcΔT

0.57kJ = 0.1(L) x 4.18(kJ/L) x ΔT

divide both sides by 0.1 x 4.18

ΔT = 1.36 degrees celsius.

 

[Munkiie]

Thank you very much!
I get it now