A student mixes a 10.0 ml sample of 1.0 m naoh(aq) with a 10.0 ml sample of 1.0 m hcl(aq) in a polystyrene container. the temperature of the solutions before mixing was 20.0°c. if the final temperature of the mixture is 26.0°c, what is the experimental value of dhrxn 

Hrxn = Q reaction / mol of reaction
mol of reaction = M * V = 10 * 1 = 10 mmol = 0.01 mol
Q water = m * C * (Tf - Ti)
= (10 + 10) (4.184) (26-20) = 502.08 J
Q reaction = - Q water = -502.08 J
Hrxn = -502.08 / (0.01) = - 50208 J = - 50.21 kJ/mol

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tutorareej tutorareej · Answer 2 · 2021-11-23T10:51:39+01:00

The value of [tex]\Delta[/tex] [tex]\rm H_r_x_n[/tex] for the experiment will be 19.278 J.

[tex]\rm H_r_x_n[/tex] can be defined as the change in enthalpy of product and reactant in the chemical reaction.

The value of [tex]\rm H_r_x_n[/tex] can be given as:

[tex]\Delta[/tex] [tex]\rm H_r_x_n[/tex] = ms[tex]\Delta[/tex]T

m = mass of reactant

s = specific heat = 4.2 g/mol [tex]\rm ^\circ C[/tex]

[tex]\Delta[/tex]T = change in temperature = 26 [tex]\rm ^\circ C[/tex] - 20 [tex]\rm ^\circ C[/tex] = 6 [tex]\rm ^\circ C[/tex]

moles = Molarity [tex]\times[/tex] Volume (L)

Moles of NaOH = 1 M [tex]\times[/tex] 0.01

Moles of NaOH = 0.01 mol

Moles of HCl = 1 M [tex]\times[/tex] 0.01

Moles of HCl = 0.01 mol

Mass of 0.01 mol HCl = 0.01 [tex]\times[/tex] 36.5 grams

Mass of 0.01 mol HCl = 0.365 grams

Mass of 0.01 mol NaOH = 0.01 [tex]\times[/tex] 40

Mass of NaOH = 0.4 grams

The total mass of reactants = 0.365 + 0.4

Mass of reactants = 0.765 grams

[tex]\Delta[/tex] [tex]\rm H_r_x_n[/tex] = 0.765 [tex]\times[/tex] 4.2 [tex]\times[/tex] 6

[tex]\Delta[/tex] [tex]\rm H_r_x_n[/tex] = 19.278 J

The value of [tex]\Delta[/tex] [tex]\rm H_r_x_n[/tex] for the experiment will be 19.278 J.

For more information about enthalpy, refer to the link:

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