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Sonja [21]
3 years ago
12

Measurements show that the enthalpy of a mixture of gaseous reactants increases by 215. kJ during a certain chemical reaction, w

hich is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -155. kJ of work is done on the mixture during the reaction.
Calculate the change in energy of the gas mixture during the reaction. Round your answer to 2 significant digits.


Is the reaction exothermic or endothermic?
Chemistry
1 answer:
Maslowich3 years ago
6 0

Answer:

1. The change in energy is 60KJ or 6.0 × 10^1 KJ

2. Endothermic reaction

Explanation:

The Enthalpy (ΔU) for the mixture is given as 215kJ and the workdone (W) on the mixture is - 155KJ. Hence, the change in the energy (ΔH) ofthe mixture is computed using the equation below:

ΔH = ΔU + w

Where,

ΔH= Change in energy

ΔU= Enthalpy change

W= workdone

Therefore ΔH is:

ΔH= 215 kJ + (-155Kj)

ΔH= 60 KJ

Therefore, the change in energy is 60KJ or 6.0 × 10^1 KJ

The value is positive so it is an endothermic reaction.

An endothermic reaction happens when the energy used to break the bonds in the reactants is higher than the energy given out when bonds are formed in the products. This means that the entire reaction takes in energy, hence there is a temperature decrease in the surroundings. Endothermic reactions cannot happen spontaneously. Work is usually done in order to get these reactions to occur. When endothermic reactions absorb energy, a temperature drop is measured by the reaction.

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ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

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Paso 2: Calcular el cociente de reacción (Qp)

Qp = pBrF × pF₂ / pBrF₃

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Paso 3: Sacar una conclusión

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