[tex]agno _3(aq) + nacl(aq) \rightarrow agcl(s)\

Ответ:

tystar84

10.01.2022

The enthalpy change during the reaction is -84 kJ/mole.

Explanation:

First we have to calculate the heat gained by the solution.

$Q=mc\times (T_{final}-T_{initial})$

where,

Q = heat gained = ?

m = mass of the solution = 100.0 g

c = specific heat = 4.2 J/^oC

$T_{final}$ = final temperature = 21.0 ^oC

$T_{initial}$ = initial temperature = 20.0 ^oC

Now put all the given values in the above formula, we get:

$Q=100.0 g\times 4.2 J/^oC\times (21.0-20.0)^oC$

Q=420 J

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat gained = 420 J

n = number of moles silver nitrate= n

$AgCl(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

Molarity of the silver nitrate solution = M = 0.100 M

Volume of the silver nitrate solution = V = 50.0 mL = 0.050 L

$n=0.100M\times 0.050 L=0.005 mol$

$\Delta H=-\frac{420 J}{0.005mole}=-84000J/mol=-84 kJ/mol$

(1 kJ = 1000 J)

Therefore, the enthalpy change during the reaction is -84 kJ/mole.

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