# calculate the enthalpy change for the neutralization please!!!?

In a coffee cup calorimeter, 100.ml of 1.0 M NaOH and 100.0 ml of 1.0M HCl are mixed. Both solutions were originally at 24.6 °C. After the reaction, the final temperature is 31.3°C. Assuming that all the solutions have a density of 1.0g/ml and a specific heat capacity of 4.18 J/g °C, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundingS or to the calorimeter.

• Total Volume of solutions after mixing: 100 + 100 = 200 mL

d = 1.0 g/mL

Total Mass of solutions after mixing: d x V = 1.0 g/mL x 200 mL = 200 g

Heat released during neutralization = m x c x dT

Q = (200 g)(4.18 J/K.C)(31.3 - 24.6)C = 5601.2 J

Heat of neutralization given in terms of kJ/mol

Since the number of moles of acid and base is the same;

n = M x V = (1.0 mol/L)(0.1 L) = 0.1 mole

Since 5601.2 J of heat is released during the neutralization of 0.1 mol, for 1 mol the amount of heat : 56012 J

dH (neutralization) = - 56.012 kJ/mol or simply - 56 kJ/mol

RE:

calculate the enthalpy change for the neutralization please!!!?

In a coffee cup calorimeter, 100.ml of 1.0 M NaOH and 100.0 ml of 1.0M HCl are mixed. Both solutions were originally at 24.6 °C. After the reaction, the final temperature is 31.3°C. Assuming that all the solutions have a density of 1.0g/ml and a specific heat capacity of 4.18 J/g °C, calculate...

• Enthalpy Change Of Neutralisation

• q = mc(delta T)

q= (200g)(4.18)(31.3 - 24.6)

q= 5601.2 j

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