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Mnenie [13.5K]
3 years ago
15

Which statement best describes general equilibrium? Equilibrium is reached when the reaction stops. There is only one set of equ

ilibrium concentrations that equals the Kc value. At equilibrium, the rate of the forward reaction is the same as the rate of the reverse reaction. At equilibrium, the total concentration of products equals the total concentration of reactants.
Chemistry
1 answer:
artcher [175]3 years ago
5 0

Answer:  The statement 'At equilibrium, the rate of the forward reaction is the same as the rate of the reverse reaction' best describes general equilibrium.

Explanation:

A chemical process in which the rate of forward reaction is equal to the rate of backward reaction.

For example, A + B \rightleftharpoons C + D

This reaction is an equilibrium process. Symbol which represents the equilibrium is '\rightleftharpoons'.

Thus, we can conclude that the statement 'At equilibrium, the rate of the forward reaction is the same as the rate of the reverse reaction' best describes general equilibrium.

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"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
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<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

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Hence, the specific heat of calorimeter is 30.68 J/g°C

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