Choose the thermochemical equation that illustrates delta H degree_f for Li_2SO_

Question

Choose the thermochemical equation that illustrates delta H degree_f for Li_2SO_4 A place of iron (mass = 25.0 g) at 395 K in a styrofoam coffee cup containing 25.0 mL of water at 298 K. Assuming that is lost to the cup or surroundings, what will the final temperature of the water be? The specific heal capacity of iron = 0.449 J/g degree C and water = 4.18 J?g degree C. Use the delta H degree _f information provided to calculate delta H degree_rxn for the following: delta H degree_f(kJ/mol) SO_2Cl_2(g) + 2 H_2O(1) rightarrow 2Hl(g) + H_2SO_4(1) delta H degree_rxn = ? Use the standard reaction enthalpies given below to determine delta H degree_rxn for the following reaction: P_4(g) + 10 Cl_2(g) rightarrow 4 PCl_5(s) delta H degree_rxn = ? How much energy is required to decompose 765 g of PCl_3, according to the reaction below? The molar mass of PCl_3 is 137.32 g.mol and may be useful. 4PCl_3(g) rightarrow P_4(s) + 6 Cl_2(g) delta H degree_rxn = +1207 kJ

Explanation / Answer

Solution :-

Q5 ) The equation shown in the option A is the correct equation because in that equation all elements are shown in their elemental states where enthalpies of formation of the reactants are 0.0 kJ therefore the enthalpy change of the reaction is same as the enthalpy change of the formation of the Li2SO4

So the correct answer is option A

Q6) mass of iron = 25 g

Temperature of iron = 398 K

Water = 25 ml = 25 g

Temperature of water = 298 K

Final temperature after mixing = ?

Iron has higher temperature so it will loose heat and water will absorb it

So the set up can be made as

-q iron = q water

- m*s*delta T = m*s*delta T

-25.0 g * 0.449 J per g C *(T2 – 398 K) = 25.0 g * 4.18 J per g C * (T2-298 K)

-11.225 T2 + 4467.55 = 104.5 T2 – 31141

4467.55 + 31141 = 104.5 T2 + 11.225 T2

35608.55 = 115.725 T2

35608.55 / 115.725 = T2

307.7 K = T2 rounded to 308 K

So the final temperature of the water is 308K

So the correct answer is 308 K that is option C

Q7) balanced reaction equation

SOCl2(g) + 2H2O(l) ----- > 2HCl (g) + H2SO4(l)

Formula to calculate the enthalpy change of reaction is as follows

Delta H rxn = sum of delta H product – sum of delta H reactant

                    =[(HCl*2)+(H2SO4*1)] –[(SOCl2*1)+(2H2O*2)]

                    = [(-92*2)+(-814*1)] –[(-364*1)+(-286*2)]

                = -62.0 kJ

So the correct answer is option C

Q8)

To get the desired equation P4(g) + 10 Cl2(g) ------ > 4PCl5(s)

We need to rearrange the given two equations

We want PCl5 on the product side

Therefore reverse equation 1 and multiply it by 4

Then add new equation to the second equation to get the desired equation

When equation 1 is reversed then its enthalpy value remains same but the sign changes

And when multiplied by 4 then delta H also gets multiplied by 4

So the delta H for the desired equation is

Delta H = (-157*4)+ (-1207) = -1835 kJ

So the correct answer is option D

Q9) 4 mol PCl3 needs +1207 kJ

So lets find the moles of the 765 g PCl3

Moles = mass / molar mass

           = 765 g / 137.3328 g per mol

           = 5.57 mol PCl3

So the delta H is calculated as

5.57 mol PCl3 * 1207 kJ / 4 mol PCl3 = 1681 kJ

So the correct answer is option C

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