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DENIUS [597]
4 years ago
5

Be sure to answer all parts. Consider the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g)ΔH = −198.2 kJ/mol How would the concentrations of S

O2, O2, and SO3 at equilibrium change if we were to (a) increase the temperature? [SO2]: The concentration increases. The concentration decreases. There is no change. [O2]: The concentration increases. The concentration decreases. There is no change. [SO3]: The concentration increases. The concentration decreases. There is no change. (b) increase the pressure and decrease the volume? [SO2]: The concentration increases. The concentration decreases. There is no change. [O2]: The concentration increases. The concentration decreases. There is no change. [SO3]: The concentration increases. The concentration decreases. There is no change. (c) add a catalyst? [SO2]: The concentration increases. The concentration decreases. There is no change. [O2]: The concentration increases. The concentration decreases. There is no change. [SO3]: The concentration increases. The concentration decreases. There is no change.
Chemistry
1 answer:
andrew-mc [135]4 years ago
4 0

Answer:

a)

[SO2]: The concentration increases

[O2]: The concentration increases.

[SO3]: The concentration decreases.

b)

[SO2]: The concentration decreases.

[O2]: The concentration decreases.

[SO3]: The concentration increases.

c)

[SO2]:There is no change.

[O2]: There is no change.

[SO3]: There is no change

Explanation:

For an exothermic reaction, increase in temperature decreases the concentration of products and increases the concentration of reactants since increase in temperature shifts the equilibrium position to the left hand side.

Increase in pressure and decrease in volume will shift the equilibrium position towards the right hand side which means more SO3 in the system.

Catalyst increases the rate of forward and reverse reaction simultaneously hence at equilibrium, the concentration of reactants and products remain unchanged.

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Answer:

6. d,  7. a

Explanation:

6. Molarity is a number of moles solute in 1 L solution.

7. 1 L solution - 2.5 mol K2CO3

 20 L             - x mol K2CO3

x  =20*2.5/1 = 50 mol K2CO3

Molar mass(KCO3) = M(K) + M(C) + 3M(O)= 39 +12 +3*16= 99 g/mol

99 g/mol *50 mol = 4950 g KCO3 Closest answer is A.

Actually KCO3 does not exist, in reality it should be K2CO3.

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4 0
3 years ago
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
A great white shark has resided at a depth where the pressure is 16 atm. If the initial volume of its swim bladder is 4 L, what
padilas [110]

Answer:

The new volume of the gas is 32L

Explanation:

P1 = 16atm

V1 = 4L

P2 = atm

V2 = ?

According to Boyle's law, the volume of a given mass of gas is inversely proportional to its volume provided the temperature remains constant.

P1 * V1 = P2 * V2

V2 = (P1 * V1) / P2

V2 = (16 * 4) / 2

V2 = 64 / 2

V2 = 32L

The new volume of the gas is 32L

6 0
4 years ago
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