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hichkok12 [17]
3 years ago
9

Consider this equilibrium reaction between carbon monoxide and hydrogen gas, occurring in a sealed flexible container. CO(g) + 3

H2g) CH4)H2)+ heat For each listed stress, choose the direction of the resulting shift. A. Shift towards reactants B. Shift towards products C. No shift occurs ▼ More H2(g) is added to the container. - COg) is removed from the container. ▼ More CH4(g) is added to the container. ▼ H2O(g) is removed from the container. - The contents of the container are heated up. - - The contents of the container are cooled down. ▼ The pressure inside the container is increased. The container is stretched to increase the volume.
Chemistry
1 answer:
wariber [46]3 years ago
6 0

Answer:

More H2(g) is added to the container : <u>Towards products.</u>

CO is removed from the container : <u>Towards reactants.</u>

More CH4(g) is added to the container : <u>Towards reactants</u>

H2O(g) is removed from the container <u>: Towards products.</u>

The contents of the container are heated up. :<u> Towards the reactants.</u>

The contents of the container are cooled down : <u>Towards the products.</u>

The pressure inside the container is increased. :<u>Towards the products</u>

The container is stretched to increase the volume: <u>Towards the reactants.</u>

Explanation: :

CO(g) + 3 H2g) → CH4(g) + H2O(g)+ heat

There is released heat, so this reaction is exothermic

If the H2 concentration is increased, the system will try to change the concentration change by shifting the balance to the right, and thus the concentration of products will increase.<u> Towards products.</u>

If the CO is removed, the system will try to change this situation by shifting the balance to the left, and thus the concentration of reactants will increase, the concentration of products will decrease. <u>Towards reactants.</u>

If the CH4 concentration is increased, the system will try to change the concentration change by shifting the balance to the left, and thus the concentration of reactants will increase. <u>Towards reactants</u>

If the H2O is removed, the system will try to change this situation by shifting the balance to the right, and thus the concentration of products will increase, the concentration of products will decrease. <u>Towards products.</u>

If the temperature is increased, the system will reduce the amount of heat released. So the balance will shift to the left. <u>Towards the reactants.</u>

This because the extra heat / energy must be used.

If the temperature is decreased, the system will produce more heat  So the balance will shift to the right. <u>Towards the products.</u>

This because more heat /energy needs to be produced to make up for the loss of heat (energy).

If the pressure is increased, the system will shift to the side with fewer moles of gas. In this case, there are 4 moles on the left and 2 moles on the right.  So the balance will shift to the right. <u>Towards the products.</u> An increase of pressure has the same effect on the equilibrium as a decrease of the volume.

If the volume is increased, this means the pressure is decreased, the system will shift to the side with most moles of gas. In this case, there are 4 moles on the left and 2 moles on the right.  So the balance will shift to the left. <u>Towards the reactants.</u> An increase of volume has the same effect on the equilibrium as a decrease of the pressure.

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3 0
3 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

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Explanation:

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The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

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lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

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3 0
3 years ago
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3 0
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alina1380 [7]

Answer:

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also the element is Cr

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