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svet-max [94.6K]
3 years ago
14

What effect does observing a substance's physical properties have on the substance?

Chemistry
1 answer:
dybincka [34]3 years ago
5 0
Depending on the testing process being used, it may affect the object by making it shape differently, scratch, it or sometimes do absolutely nothing.  hope this helps:)
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Hazardous waste, potential reactor accident, and depletes natural waters
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Is the energy change accompanying the synthesis of a compound from its elements in their standard state?
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The synthesis of a compound from its elements in their standard state, accompanied by a change in energy that is known as standard enthalpy of formation of a compound. The standard state of a substance is the most stable physical state of the compound that exist at STP that is at 1 atm, 273 K temperature. So the answer is yes, the energy change accompanying the synthesis of a compound from its elements in their standard state.

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For the reaction co(g)+h2o(g)⇌h2(g)+co2(g), k= 4.24 at 800 k what can be said about this reaction at this temperature?
Amanda [17]

The question has missing information.

The options are given below.

a.The equilibrium lies far to the right

b. The reaction will proceed very slowly

c. The reaction contains significant amounts of products and reactants at equilibrium.

d. The equilibrium lies far to the left

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.

Therefore option C is the correct option.

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4 years ago
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