Chemistry, 06.05.2020 02:19 becddjr

When 22.7 mL of 0.500 M H2SO4 is added to 22.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH of this reaction. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)___kJ/mol H2O

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Answer from: bigdramm409

Answer : The enthalpy of neutralization is, -55.8 KJ/mole

Explanation :

First we have to calculate the moles of H_2SO_4 and KOH.

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.500mole/L\times 0.0227L=0.01135mole$

$\text{Moles of KOH}=\text{Concentration of KOH}\times \text{Volume of solution}=1.00mole/L\times 0.0227L=0.0227mole$

Now we have to calculate the limiting reagent.

The balanced chemical reaction will be,

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of H_2SO_4 react to give 2 moles of H_2O

So, 0.01135 mole of H_2SO_4 react to give $2\times 0.01135=0.0227$ moles of H_2O

and,

As, 2 moles of KOH react to give 2 moles of H_2O

As, 0.0227 moles of KOH react to give 0.0227 moles of H_2O

From this we conclude that they can form the same amount of product. This means that both will consume completely in the reaction.

Thus, the number of neutralized moles = 0.0227 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 22.7ml+22.7ml=45.4ml

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 45.4ml=45.4g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

= specific heat of water = 4.184J/g^oC

m = mass of water = 45.4 g

$T_{final}$ = final temperature of water = 30.17^oC

$T_{initial}$ = initial temperature of water = 23.50^oC

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

$q=45.4g\times 4.184J/g^oC\times (30.17-23.50)^oC$

q=1266.99J=1.267kJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -1.267 KJ

n = number of moles used in neutralization = 0.0227 mole

$\Delta H=\frac{-1.267KJ}{0.0227mole}=-55.8KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, -55.8 KJ/mole

Answer from: Quest

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

$^{222}_{86} rn+ ^{4}_{2} he$ →

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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When 22.7 mL of 0.500 M H2SO4 is added to 22.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50...

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