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3 years ago

7

Measurements show that enthalpy of a mixture of gaseous reactants decreases by 228. kJ during a certain chemical reaction, which

is carried out at a constant pressure. Furthermore, by carefully monitoring the volume change it is determined that -55kJ of work is done on the mixture during the reaction.
Calculate the change in energy of the gas mixture during the reaction.

Is the reaction exothermic or endothermic?

2 answers:

Tcecarenko [31]3 years ago

7 0

Total change in energy = Heat produced by system + work done on the system
= 228 - 55
= 173 kJ

Because this energy is released as heat, the reaction is exothermic.

Montano1993 [528]3 years ago

7 0

Answer:

-283 KJ, exothermic

Explanation:

We are given that

Enthalpy of a mixture of a gases reactants

Change in enthalpy= $\Delat H$ =-228 KJ

Where negative sign represents the enthalpy decreases.

Pressure=Constant.

Work don=w=-55 KJ

We have to calculate the change in energy of the gas mixture during the reaction.

At constant pressure, the change in enthalpy

$\Delta H=\Delta U+P\Delta V$

Where w= $-P\Delta V$

Where P=Constant

$\Delta V=$ Change in volume

$\Delta U=$ Change in energy

$\Delta U=\Delta H-P\Delta V$

$\Delta U=\Delta H+w$

Substitute the values then we get

$\Delta U=-228-55=-283$ KJ

Hence, the change in energy of the gas mixture during the reaction=-283 KJ

Change in enthalpy is negative it means the reaction is exothermic because the energy is evolved.

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4 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

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Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

A)

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

B)

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

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4 years ago