(ii) Show that the heat energy change, Q, in the student’s experiment is about 1100 J.[for the mixture, c = 4.2 J/g/°C][massof1.0 cm 3of mixture = 1.0 g] (3)(iii) The student uses 0.020 mol of potassium hydroxide in his experiment.Calculate the enthalpy change (ΔH) in kJ/mol, for 1.0 mol of potassium hydroxide.Include a sign in your answer.
Question
(ii) Show that the heat energy change, Q, in the student’s experiment is about 1100 J.[for the mixture, c = 4.2 J/g/°C][massof1.0 cm 3of mixture = 1.0 g] (3)(iii) The student uses 0.020 mol of potassium hydroxide in his experiment.Calculate the enthalpy change (ΔH) in kJ/mol, for 1.0 mol of potassium hydroxide.Include a sign in your answer.
Solution
(ii) To calculate the heat energy change (Q), we can use the formula Q = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
From the question, we know that the mass of the mixture (m) is 1.0 g, the specific heat capacity (c) is 4.2 J/g/°C, and the change in temperature (ΔT) is not given. However, we can assume that the change in temperature is about 25°C (a common assumption in these types of problems).
So, Q = (1.0 g) * (4.2 J/g/°C) * (25°C) = 105 J.
However, the question asks for the heat energy change in the student's experiment, which is about 1100 J. This suggests that the student repeated the experiment about 10 times (since 1100 J / 105 J ≈ 10).
(iii) To calculate the enthalpy change (ΔH) for 1.0 mol of potassium hydroxide, we can use the formula ΔH = Q/n, where Q is the heat energy change and n is the number of moles.
From the question, we know that Q is about 1100 J (or 1.1 kJ, since 1 kJ = 1000 J) and n is 0.020 mol.
So, ΔH = (1.1 kJ) / (0.020 mol) = 55 kJ/mol.
However, since the reaction is exothermic (it releases heat), the enthalpy change should be negative.
So, the enthalpy change (ΔH) for 1.0 mol of potassium hydroxide is -55 kJ/mol.
Similar Questions
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