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n200080 [17]
4 years ago
14

Which of the following phase diagrams represents how a catalyst is able to change the rate of a reaction? (2 points) A regular e

xothermic potential energy diagram is shown, with a dotted line representing how the potential energy diagram changes with the use of a catalyst. The dotted line shows a new potential energy diagram with two potential energy hills, instead of the one hill in the regular potential energy diagram. A regular exothermic potential energy diagram is shown, with a dotted line representing how the potential energy diagram changes with the use of a catalyst. The dotted line shows a new potential energy diagram with a taller activation energy hill than that in the original potential energy diagram. A regular exothermic potential energy diagram is shown, with a dotted line representing how the potential energy diagram changes with the use of a catalyst. The dotted line shows a new potential energy diagram with two activation energy hills, the second taller than the first, instead of the one hill in the regular potential energy diagram. A regular exothermic potential energy diagram is shown, with a dotted line representing how the potential energy diagram changes with the use of a catalyst. The dotted line shows a new potential energy diagram with a shorter activation energy hill than that in the original potential energy diagram.

Chemistry
1 answer:
OverLord2011 [107]4 years ago
7 0

Answer: The dotted line shows a new potential energy diagram with a shorter activation energy hill than that in the original potential energy diagram.

Explanation:

Activation energy is the extra amount of energy required by the reactants to cross the energy barrier to get converted into products.

When a catalyst is added, it lowers down the activation energy which is shown by a dotted line. As now less energy is required,  more of reactants can cross the energy barrier and get converted to products and thus increase the rate of reaction.

A catalyst increases the rate of reaction for both exothermic and endothermic reactions.

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The question has missing information.

The options are given below.

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Answer : The correct answer is option c: The reaction contains significant amounts of products and reactants at equilibrium.

Explanation :

Option a is not correct because for equilibrium to lie far to the right, we need a large value of keq ( Keq >>> 1 ), but the keq value is small.

option b is also not correct because in order to predict the speed of the reaction, we need more information. The speed of the reaction cannot be predicted using Keq value.

Option d is not correct because for Keq to lie far to the left, it has to be very very small. ( Keq <<< 1)

The value of equilibrium constant is given as 4.24. This value is very near to 1.

Equilibrium constant for this reaction is calculated as

Keq = \frac{[H_{2}] [CO_{2}]}{[H_{2}O][CO]}

When Keq = 1, [H_{2}] [CO_{2}  ] = [CO] [H_{2}O]

That means concentrations of products and reactants are equal.

Since the keq value for our reaction is very close to 1, we can say that , there are significant amounts of reactants and products at equilibrium.

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