Answer:
100 ml of a 0.300 m solution of agno3 reacts with 100 ml of a 0.300 m solution of hcl in a coffee-cup calorimeter and the temperature rises from 21.80 °c to 23.20 °c. Assuming the density and specific heat of the resulting solution is 1.00 g/ml and 4.18 j/g ∙ °c respectfully, what is the ΔH°rxn?
39.013 kJ/mol.
Explanation:
AgNO3(aq) + HCl(aq) --------------> AgCl(s) + HNO3(aq)
We can calculate the amount of heat (Q) released from the solution using the relation:
Q = m.c.ΔT,
Where, Q is the amount of heat released from the solution (Q = ??? J).
m is the mass of the solution (m of the solution = density of the solution x volume of the solution = (1.0 g/mL)(200 mL) = 200 g.
c is the specific heat capacity of the solution (c = 4.18 J/g∙°C).
ΔT is the difference in the T (ΔT = final temperature - initial temperature = 23.20 °C - 21.80 °C = 1.4 °C).
∴ Q = m.c.ΔT = (200 g)(4.18 J/g∙°C)(1.4 °C) = 1170.4 J.
∵ ΔH°rxn = Qrxn/(no. of moles of AgNO₃).
Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.
M = (no. of moles of AgNO₃)/(Volume of the solution (L)).
∴ no. of moles of AgNO₃
= (M)(Volume of the solution (L))
= (0.3 M)(0.1 L) = 0.03 mol.
∴ ΔH°rxn
= Qrxn/(no. of moles of AgNO₃)
= (1170.4 J)/(0.03 mol)
= 39013.33 J/mol
= 39.013 kJ/mol.
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= 0.10![\frac{[H^+][0.05]}{[0.1]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5B0.05%5D%7D%7B%5B0.1%5D%7D%20)
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] = -log(2.513*10^-5)= 4.59.