The enthalpy change of the precipitation reaction is 84 kJ/mole
Why?
The chemical equation for the reaction is
AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)
To find the enthalpy change we need to apply the following equation
[tex]\Delta H =\frac{Q}{n}[/tex]
To find the heat (Q):
[tex]Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J[/tex]
Now, to find the number of moles that react (n):
[tex]n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles[/tex]
Having these two values we can plug in the first equation:
[tex]\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole[/tex]
Have a nice day!
The change in enthalpy for the reaction has been 84 kJ/mol.
The chemical equation for the reaction will be:
[tex]\rm AgNO_3\;+\;NaCl\;\rightarrow\;AgCl\;+\;NaNO_3[/tex]
The enthalpy change can be given as the ratio of heat to moles.
Moles of Silver nitrate = molarity [tex]\times[/tex] volume
Moles of Silver nitrate = 0.1 [tex]\times[/tex] 0.05 L
Moles of Silver nitrate = 0.005 moles.
Heat can be given by:
Q = mc[tex]\Delta[/tex]T
m = mass = 100 grams
c = specific heat capacity = 4.2 J/g[tex]\rm ^\circ C[/tex]
[tex]\Delta[/tex]T = change in temperature = 21 [tex]\rm ^\circ C[/tex] - 20 [tex]\rm ^\circ C[/tex] = 1 [tex]\rm ^\circ C[/tex]
Q = 100 [tex]\times[/tex] 4.2 [tex]\times[/tex] 1
Q = 420 J
The enthalpy change ([tex]\Delta[/tex]H) can be given by:
[tex]\Delta[/tex]H = [tex]\rm \dfrac{Heat}{moles}[/tex]
[tex]\Delta[/tex]H = [tex]\rm \dfrac{420}{0.005}[/tex]
[tex]\Delta[/tex]H = 84000 J/mol
[tex]\Delta[/tex]H = 84 kJ/mol
The change in enthalpy for the reaction has been 84 kJ/mol.
For more information about enthalpy change, refer to the link:
https://brainly.com/question/1477087