Chemistry: When 4.98 g of NaOH was dissolved in 52.79 g of water in a calorimeter at 23.7 oC, the temperature of the solution went up to 50.1 oC. What is the enthalpy change in kJ/mole of sodium hydroxide? Assume the specific heat of the mixture is the same as water.

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When 4.98 g of NaOH was dissolved in 52.79 g of

Ответ:

larkinc2946

24.01.2022

Chemistry

Change in enthalpy is -51.072 kJ/mol.

Explanation:

Molar mass of NaOH = 39.997 g/mol

So, 4.98 g of NaOH = $\frac{4.98}{39.997}$ moles of NaOH = 0.125 moles of NaOH

Total mass of solution = (4.98+52.79) g = 57.77 g

Heat consumed by solution = $(m_{solution}\times C_{solution}\times \Delta T_{solution})$ , where m is mass C is specific heat and $\Delta T$ is change in temperature.

So, heat consumed by solution = $[57.77g\times 4.186\frac{J}{g.^{0}\textrm{C}}\times (50.1-23.7)^{0}\textrm{C}]$

= 6384 J

It is an exothermic process as temperature increases during dissolution. Hence change in enthalpy should be negative.

Change in enthalpy = -(heat consumed by solution)/(no. of moles of NaOH)

= $-\frac{6384J}{0.125mol}=-51072J/mol=-51.072kJ/mol$

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When 4.98 g of NaOH was dissolved in 52.79 g of water in a calorimeter at 23.7 oC, the temperature of the solution went up to 50.1 oC. What is the enthalpy change in kJ/mole of sodium hydroxide? Assume the specific heat of the mixture is the same as water.

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