Chemistry-concentration

2013-11-20 11:58 pm

A student performing a calorimeter experiment combined 100.0 mL of 0.50 M
HCl and 100.0 mL of 0.50 M NaOH in a StyrofoamTM cup calorimeter. Both
solutions were initially at 20.0 C, but when the two were mixed, the temperature
rose to 23.2 C.

(a) Suppose the experiment is repeated in the same calorimeter but this time
using 200.0 mL of 0.50 M HCl and 200.0 mL of 0.50 M NaOH. Will the ∆T
observed be greater than, less than, or equal to that in the first experiment,
and why?

(b) Suppose the experiment is repeated once again in the same calorimeter, this
time using 100.0 mL of 1.00 M HCl and 100.0 mL of 1.00 M NaOH. Will the
∆T observed be greater than, less than, or equal to that in the first experiment,
and why?

回答 (1)

micatkie

2013-11-21 3:18 am

✔ 最佳答案

E = m c ΔT

Assuming there is no heat loss to the surroundings.
E (the energy given off) is directly proportional to the number of moles ofH2O formed in neutralization.
m (the mass of solution to absorb heat) is directly proportional to the totalvolume of final solution.
c (the specific heat capacity of the final solution) is a constant.

When (100.0 mL of 0.50 M HCl) + (100.0 mL of 0.50 M NaOH) :
Number of moles of H2O formed = 0.50 x (100.0/1000) = 0.05 mol
Volume of solution to absorb heat = 100 + 100 = 200 mL

(a)
ΔT is equal to that in the firstexperiment.

When (100.0 mL of 0.50 M HCl) + (100.0 mL of 0.50 M NaOH) :
Number of moles of H2O formed = 0.50 x (200.0/1000) = 0.10 mol
Volume of solution to absorb heat = 200 + 200 = 400 mL
Hence, the heat given off and the volume of solution to absorb heat are both doubled. As a result, ΔT is unchanged.

(b)
ΔT is greater than that in the firstexperiment.

When (100.0 mL of 1.00 M HCl) + (100.0 mL of 1.00 M NaOH) :
Number of moles of H2O formed = 1.00 x (100.0/1000) = 0.10 mol
Volume of solution to absorb heat = 100 + 100 = 200 mL
Hence, the heat given off is double, but the volume of solution to absorb heat remainsunchanged. As a result, ΔT becomesgreater.

參考: micatkie

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