Chemistry, 02.02.2020 20:57 b00n4l1f3

Which statement about enthalpy is true?

the enthalpy of formation for a pure element in its standard state is always positive.
the enthalpy of formation for a pure element in its standard state is always negative.
enthalpy is a state function because its change depends only on initial and final conditions.
enthalpy is not a state function because its change depends on the identities of the reactants and products.

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Answer from: ThunderThighsM8

Enthalpy is a state function because its change depends only on initial and final conditions.

Answer from: pgfrkypory2107

Answer : Option C) Enthalpy is a state function because its change depends only on initial and final conditions.

Explanation : Enthalpy can be defined as a thermodynamic quantity which is equivalent to the total heat content of a system. It is equal to the internal energy of the system(U) plus the product of pressure and volume (PV).

E = U + PV ;

It is a state function as it only depends on the initial and final conditions, and not on the path taken to reach these reaction conditions.

Therefore, the integral of state functions can be taken using only two values: the final and initial values.

Answer from: Quest

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Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Which statement about enthalpy is true?

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