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RoseWind [281]
2 years ago
8

a sample of sulfur dioxide occupies a volume of 652 mL at 40.0 C and 0.75 atm. What volume will the sulfur dioxide occupy at STP

?
Chemistry
1 answer:
sesenic [268]2 years ago
5 0

The volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles. Details about volume can be found below.

<h3>How to calculate volume?</h3>

The volume of a gas can be calculated using the following formula:

PV = nRT

  • P = pressure
  • V = volume
  • n = number of moles
  • R = gas law constant
  • T = temperature

0.75 × 0.652 = n × 0.0821 × 313

0.489 = 25.69n

n = 0.489/25.69

n = 0.019moles

Therefore, the volume that sulfur dioxide will occupy with a volume of 652 mL at 40.0°C and 0.75 atm is 0.019moles.

Learn more about volume at: brainly.com/question/1578538

#SPJ1

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Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

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K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

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