Answer:
[tex]\Delta H_{rxn}=162940 J/mol[/tex]
Explanation:
To calculate the heat of reaction we need to determine:
Moles of zinc (Zn)
It says that the 0.103 g reacted completely, knowing that the Zn molecular weight is 65.4 g/mol:
[tex]n_{Zn}=\frac{0.103g}{65.4 g/mol}=1.57*10^{-3}mol[/tex]
Heat released by the reaction
The solution goes from 22.5 °C to 23.7°C:
[tex]Q=Cp*\rho *V*(T_f - T_i)[/tex]
[tex]Q=4.18 \frac{J}{g* ^{\circ}C}*1.02 g/mL*50mL*(23.7-22.5)^{\circ}C[/tex]
[tex]Q=255.8 J[/tex]
Heat of reaction
[tex]\Delta H_{rxn}=\frac{-Q}{n_{Zn}}[/tex]
[tex]\Delta H_{rxn}=\frac{-255.8 J}{1.57*10^-3}}=162940 J/mol[/tex]
The Delta [tex]H_rxn[/tex] is the enthalpy of the reaction. The Delta [tex]H_rxn[/tex] per mole of Zn is 162940 j/mol.
The enthalpy of reaction ([tex]H_rxn[/tex]) is the difference in enthalpy between total products and total reactants in a chemical reaction.
To calculate the enthalpy of the reaction
We will find moles of zinc
The heat of the reaction
Find the moles of zinc
[tex]\rm Number\;of \;moles= \dfrac{mass}{molar\;mass}\\\\Number\;of \;moles= \dfrac{ 0.103 g}{65.4\;g/mol} = 1.57\times 10^-^3\;mol[/tex]
The heat released
[tex]\rm Q= C_p \times p\times V \times (T_f-T_i)\\\\Q= 4.18\;j/g\times\circ C \times 1.02\;g/ml \times 50\;ml \times(23.7-22.5)\circ C[/tex]
[tex]Q= 255.8\;J[/tex]
The heat of reaction is
[tex]\Delta H_rxn = \dfrac{-Q}{n_Zn} \\\\\\\Delta H_rxn = \dfrac{-255.8\;J}{1.57\times 10^-^3}= 162940 \;J /mol[/tex]
Thus, the enthalpy of the reaction is 162940 j/mol.
Learn more about enthalpy, here:
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