n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 degrees

Question

In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH is mixed with 100.0 mL of 1.0 M HCl. Both solutions were initially at 24.6 degrees Celsius. After the reaction, the final temperature of the mixture is measured at 31.3 degrees Celsius. Assuming that the density of the solution is 1.0 g/cm³ and the specific heat capacity is 4.18 J/(°C·g), calculate the enthalpy change for the neutralization of HCl by NaOH. Please assume that no heat is lost to the surroundings or to the calorimeter.

Cornelius Franecki · Accepted Answer

56.0kJ/molExplanation:The reaction of NaOH with HCl is:NaOH + HCl → H₂O + NaCl + ΔH Where ΔH is the heat of reaction that is released per mole of reactants,The moles that reacts are:100mL = 0.1L * (1mol / L) = 0.1 moles reactsTo find the heat released in the coffee cup calorimeter, you use the equation:Q = m×ΔT×C Where Q is heat released,m is mass of the solutionΔT is change in temperature (Final temperature - Initial temperature)C is specific heat of the solution (4.18J/g°C)Mass of the solution is:100mL + 100mL = 200mLDensity of the solution is 1.0g/mL. The mass is 200gChange in temperature is 31.3°C - 24.6°C = 6.7°CReplacing:Q = m×ΔT×CQ = 200g×6.7°C×4.18J/g°CQ = 5601.2JThis is the heat released per 0.1mol. The heat released per mole (Enthalpy change for the neutralization of HCl by NaOH is:5601.2J / 0.1 moles = 56012J / mol = 56.0kJ/mol

Devonte Schimmel Sr. · Answer

Explanation:mass of the solution = volume x density = 200 x 1 = 200 gm heat absorbed = m x s x Δ t , s is specific heat , Δt is rise in temperature = 200 x 4.18 x ( 31.3 - 24.6 )= 5601 J . This is the enthalpy change required.

Anonymous

n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 degrees

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