Question

Zinc metal reacts with HCl according to the balanced equation:Zn + 2HCl à ZnCl2 + H2When 0.103 g of Zn is combined with enough HCl to make 50 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 C to 23.7 C. Find the heat of the reaction. (ans: -160 kJ/mol)

Answer 1

-163kJ/mol

Explanations:

Given the reaction between the zinc metal reacts with HCl according to the balanced equation:

`Zn+2\text{HCl}\rightarrow\text{ZnCl}_2+H_2`

The required heat of reaction will be calculated using the formula below:

`\triangle H_{\text{rxn}}=-(Q)/(n_(Zn))`

where:

• Q is the ,heat produced

,

• nZn is the, number of moles that reacted

The formula for calculating the quantity of heat produced is expressed as:

`Q=mc\triangle\theta`

where:

• m is the ,mass, of the ,Zinc metal

,

• c is the ,specific heat capacity, of zinc

,

• △θ is th,e change in temperature

Get the mass of zinc

`\begin{gathered} \text{mass}=\text{ }density*\text{volume} \\ m=\rho* v \end{gathered}`

The quantity of heat becomes:

`Q=\rho\cdot v\cdot C\triangle\theta`

Substitute the given parameters to have:

`\begin{gathered} Q=\frac{1.02\cancel{g}}{\cancel{mL}}*50\cancel{mL}*\frac{4.18J}{\cancel{g^oC^{}}}*(23.7-22.5)\cancel{^oC} \\ Q=1.02*50*4.18*1.2 \\ Q=255.8Joules \end{gathered}`

Next is to get the number of moles of Zinc that reacted (nZn)

`\begin{gathered} nZ_n=\frac{Mass}{\text{Molar mass}} \\ nZ_n=\frac{0.103g}{65.4g\text{/mol}} \\ nZ_n=0.00157mole \end{gathered}`

Get the required heat of reaction of Zinc:

`\begin{gathered} \triangle H_{\text{rxn}}=-(Q)/(nZ_n) \\ \triangle H_{\text{rxn}}=-(255.8Joules)/(0.00157moles) \\ \triangle H_{\text{rxn}}=-162,929.93J\text{/mol} \\ \triangle H_{\text{rxn}}\approx-163kJ\text{/mol} \end{gathered}`

Hence the heat of the reaction of Zinc is approximately -163kJ/mol