(a) A 102-g sample of water is placed in an insulated container and allowed to come to room temperature at 21°C. To heat the water sample to 41°C, how much heat must you add to it?
kJ
(b) Consider the hypothetical reaction, being run in an insulated container that contains 102 g of solution.
2 X(aq) + Y(l) → X2Y(aq)
If the temperature of the solution changes from 21°C to 31°C, how much heat does the chemical reaction produce? (You can assume that this solution is so dilute that it has the same heat capacity as pure water.)
kJ
How does this answer compare with that in part (a)? (Enter your answer using only a number or fraction.)
This is of the value of q in part (a).
(c) If you wanted the temperature of 102 g of this solution to increase from 21°C to 51°C, how much heat would you have to add to it? (Try to answer this question without using a formula.)
kJ
(d) If you had added 0.02 mol of X and 0.01 mol of Y to form the solution in part b, how many moles of X and Y would you need to bring about the temperature change described in part c.
X mol Y mol
(e) Judging on the basis of your answers so far, what is the enthalpy of the reaction 2 X(aq) + Y(l) → X2Y(aq)?
kJ
(a)
q = mass x specific heat x delta T
q = 102 x 4.184 x (41-21)
(a) To calculate the amount of heat required to heat the water sample from 21°C to 41°C, you need to use the formula:
q = m * c * ΔT
Where:
q = heat
m = mass of the water sample (102 g)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (41°C - 21°C = 20°C)
To convert from joules (J) to kilojoules (kJ), divide the result by 1000.
q = (102 g) * (4.184 J/g°C) * (20°C) / 1000 = 86.54 kJ
Therefore, you need to add approximately 86.54 kJ of heat to the water sample.
(b) Since the solution is so dilute that it has the same heat capacity as pure water, you can use the same formula as in part (a) to calculate the amount of heat produced by the chemical reaction:
q = m * c * ΔT
Where:
q = heat
m = mass of the solution (102 g)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (31°C - 21°C = 10°C)
q = (102 g) * (4.184 J/g°C) * (10°C) / 1000 = 4.32 kJ
Therefore, the chemical reaction produces approximately 4.32 kJ of heat.
To compare this with part (a), divide the heat produced by the chemical reaction (4.32 kJ) by the amount of heat required to heat the water (86.54 kJ):
4.32 kJ / 86.54 kJ = 0.05
So, the amount of heat produced by the chemical reaction is 0.05 (or 1/20) of the amount of heat required to heat the water.
(c) Since the specific heat capacity of the solution is the same as pure water, you can use the same formula as in part (a) to calculate the amount of heat required to increase the temperature from 21°C to 51°C:
q = m * c * ΔT
Where:
q = heat
m = mass of the solution (102 g)
c = specific heat capacity of water (4.184 J/g°C)
ΔT = change in temperature (51°C - 21°C = 30°C)
q = (102 g) * (4.184 J/g°C) * (30°C) / 1000 = 125.95 kJ
Therefore, you need to add approximately 125.95 kJ of heat to the solution.
(d) Without using a formula, we can reason that if the temperature change is the same (30°C) and the specific heat capacity is the same (equal to that of pure water), the amount of heat required will be the same. Therefore, you would need to add the same amount of heat as in part (c), which is approximately 125.95 kJ.
Since the reaction stoichiometry is 2 X(aq) + Y(l) → X2Y(aq), you would need twice as many moles of X compared to Y to bring about the temperature change described. Therefore, you would need 2 * 0.02 mol of X and 0.01 mol of Y.
(e) Since the enthalpy change (ΔH) is equal to the heat (q) produced or absorbed in a reaction, based on the results so far:
In part (a), the heat required to heat the water was 86.54 kJ.
In part (b), the reaction produced 4.32 kJ of heat.
The change in enthalpy (ΔH) for the reaction can be estimated by calculating the difference between the heat required (part a) and the heat produced (part b):
ΔH = heat required - heat produced
ΔH = 86.54 kJ - 4.32 kJ = 82.22 kJ
Therefore, the enthalpy of the reaction 2 X(aq) + Y(l) → X2Y(aq) is approximately 82.22 kJ.