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3 years ago

Given the standard entropy values in the table, what is the value of `Delta"Sº"_"reaction"` for the following reaction?

Chemistry

2 answers:

Verizon [17]3 years ago

8 0

Answer:

$\Delta S^o=-149\frac{J}{mol*K}$

Explanation:

Hello,

In this case, we define the standard change in the entropy for a chemical reaction as:

$\Delta S^o=\Sigma \nu _iS_i^o$

Whereas $\nu _i$ is the stoichiometric coefficient of the species i which is negative for products and positive for reactants and $S_i^o$ is the standard entropy for the species i in J/mol*K. Thus, we have:

$\Delta S^o=(-2)(53} )+(-3)(205)+(2)(248)+(2)(38)\\\Delta S^o=-149\frac{J}{mol*K}$

Best regards.

Fantom [35]3 years ago

4 0

Answer:

entropy of reaction = 572-718= -146J/mol-k.

Explanation:

entropy of a reaction = entropy of product - entropy of reactant.

entropy of product = 2×entropy of SO2+2×entropy of NiO

= $2\times 248+2\times 38$

=572 J/mol-k

entropy of reactant = 2×entropy if NiS+3×entropy of O2

= $2\times 53+3\times 205$

=718 J/mol-k

therefore entropy of reaction = 572-718= -146J/mol-k.

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Calculate the freezing point and boiling point of a solution containing 8.15 g of ethylene glycol (C2H6O2) in 96.3 mL of ethanol

pishuonlain [190]

Answer: The freezing point of solution is -117.54°C and the boiling point of solution is 80.48°C

Explanation:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.789 g/mL

Volume of ethanol = 96.3 mL

Putting values in above equation, we get:

$0.789g/mL=\frac{\text{Mass of ethanol}}{96.3mL}\\\\\text{Mass of ethanol}=(0.789g/mL\times 96.3mL)=75.98g$

Calculating the freezing point:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

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Freezing point of pure solution = -114.1 °C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.99°C/m

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$-114.1-\text{Freezing point of solution}=1\times 1.99^oC/m\times \frac{8.15\times 1000}{62g/mol\times 75.98}\\\\\text{Freezing point of solution}=-117.54^oC$

Hence, the freezing point of solution is -117.54°C

Calculating the boiling point:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

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Boiling point of pure solution = 78.4°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 1.20°C/m.g

$m_{solute}$ = Given mass of solute (ethylene glycol) = 8.15 g

$M_{solute}$ = Molar mass of solute (ethylene glycol) = 62 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 75.98 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-78.4=1\times 1.20^oC/m\times \frac{8.15\times 1000}{62\times 75.98}\\\\\text{Boiling point of solution}=80.48^oC$

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