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3 years ago

What describes the change in Gibbs free energy (∆G) of a spontaneous reaction?

Chemistry

1 answer:

alexandr1967 [171]3 years ago

3 0

Answer: The correct answer is Option A.

Explanation:

Gibbs Free energy is defined as amount of useful work that can be done in the system. It is equal to the enthalpy of the reaction minus the product of entropy of the reaction and absolute temperature.

Mathematically,

$\Delta G=\Delta H-T\Delta S$

Where,

$\Delta G$ = Gibbs free energy

$\Delta H$ = Enthalpy of the reaction

$\Delta S$ = Entropy of the reaction

T = Absolute temperature

Sign Convention for Gibbs Free energy:

If $\Delta G$ , it is considered as a Spontaneous reaction.

If $\Delta G>0$ , it is considered as a Non-Spontaneous reaction.

If $\Delta G=0$ , reaction is in equilibrium.

Hence, the correct answer is Option A.

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If a 2.75 L gas sample is held at a constant temperature, and its pressure is changed from 750.0 torr to 360.0 torr, what will t

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3 years ago

You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft

olya-2409 [2.1K]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

$Mass=Density\times Volume$

Volume of solution = Volume of HCl + Volume of $Ba(OH)_2$

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

$Mass=1g/mL\times 93.1mL=93.1g$

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat released = ?

m = mass = 93.1 g

$C_p$ = specific heat capacity of water = $4.184J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = $29.83^oC$

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

$Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC$

$Q=1881.43J=1.88kJ$ (1 kJ = 1000 J)

Now we have to calculate the moles of $Ba(OH)_2$ and HCl.

$\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}$

$\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}$

$\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $Ba(OH)_2$ react with 2 mole of $HCl$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $HCl$

From this we conclude that, $HCl$ is an excess reagent because the given moles are greater than the required moles and $Ba(OH)_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$