Question

When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of the MnMn reacts, raising the temperature of the solution from 23.1 ∘C∘C to 28.9 ∘C∘C. Find ΔHrxnΔHrxn for the reaction as written. (Assume that the specific heat capacity of the solution is 4.18 J/g∘CJ/g∘C and the density is 1.00 g/mLg/mL.)

Answer 1

The enthalpy change during the reaction is -199. kJ/mol.

Step-by-step explanation:

`Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)`

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

`m=1.00 g/mL* 100.0 mL = 100 g`

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

`q=m* c* (T_(final)-T_(initial))`

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat =
`4.18 J/^oC`

`T_(final)` = final temperature =
`23.1^oC`

`T_(initial)` = initial temperature =
`28.9^oC`

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

`q=100 g * 4.18 J/^oC* (28.9-23.1)^oC`

`q=2,242.4 J=2.242 kJ`

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

`\Delta H=-(q)/(n)`

where,

`\Delta H` = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose =
`\frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=(0.620 g)/(54.94 g/mol)=0.0113 mol`

`\Delta H=-(2.242 kJ)/(0.0113 mol )=-199. kJ/mol`

Therefore, the enthalpy change during the reaction is -199. kJ/mol.