The value of enthalpy for the following reaction is -52.0 kJ/mol NH3 (aq) + HCl
ID: 515126 • Letter: T
Question
The value of enthalpy for the following reaction is -52.0 kJ/mol NH3 (aq) + HCl (aq) ---> NH4Cl(aq) Estimate the increase in temperature that will occur when 100.0mLs of .50M ammonia in a styrofoam container is neutralized by 300 mLs of a solution containing excess HCl. The value of enthalpy for the following reaction is -52.0 kJ/mol NH3 (aq) + HCl (aq) ---> NH4Cl(aq) Estimate the increase in temperature that will occur when 100.0mLs of .50M ammonia in a styrofoam container is neutralized by 300 mLs of a solution containing excess HCl. NH3 (aq) + HCl (aq) ---> NH4Cl(aq) Estimate the increase in temperature that will occur when 100.0mLs of .50M ammonia in a styrofoam container is neutralized by 300 mLs of a solution containing excess HCl.Explanation / Answer
No of mol of NH3 =( 0.5mol /1000)× 100ml = 0.05mol
Heat of neutralization = -52KJ/mol = -52000J/mol
For 0.05 mol Heat released = (-52000/1)×0.05= - 2600J
Heat capacity of water = 4.18 J/gmol
Mass of water = 400g
q = m× T × C
2600 J = 400g × T × 4.18(J/g K )
T = 1.55
Therefore, Increase in temperature is 1.55
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