Homework Answers
The given equilibrium can be written as,
2 NO2(g) <======> N2O4(g) ; dH = -ve
So the reaction is exothermic.
1) A
When the temperature is increased the equilibrium shifts in backward direction, so reactants conc will be higher than products concentration. So, equilibrium constant value decreases.
K = [Products] / [Reactants]
2) B
When the temperature is increased the equilibrium shifts in backward direction, so equilibrium concentration of NO2 will increase.
3) A
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Dinitrogen tetraoxide and nitrogen dioxide are two gases that exist in equilibrium at a range of temperatures. NO2 is a reddish brown gas while N204 is colorless. At High Temperature the red color is strong. || At Low Temperature the gas has less color. If we represent the equilibrium as: 2 NO2(g) = N204(9) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the...
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please help 1. Dinitrogen tetraoxide (colorless gas) is converted to nitrogen dioxide (dark reddish brown gas)...
please help
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The second question is either true or false.
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Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which...
Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which is a reddish brown gas. N204(9) 2 NO2(g) Three experiments were run starting with different initial amounts of N204(g) ([N2040 in the table). The systems were allowed to reach equilibrium and the concentrations for each gas were measured (in units of moles/Liter). For each of the boxes below, calculate the ratios indicated by the column heading. This will show the mathematical relationship that exists...
Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which...
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At High Temperature the red color is strong. At Low Temperature the gas has less color....
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When heated, colorless dinitrogen tetraoxide, N204(8), decomposes into red-brown nitrogen dioxide, NO2(g), according to the following...
When heated, colorless dinitrogen tetraoxide, N204(8), decomposes into red-brown nitrogen dioxide, NO2(g), according to the following reaction: N204(g) 42 NO2 (g) Suppose that 2.00 mol of N204(8) was placed into an empty 5.00-L flask and heated to 407 K. When equilibrium was attained, the concentration of red-brown NO2(g) was found to be 0.525 M. Calculate the equilibrium constant, K., for this reaction at 407 K. 0.500 O 0.525 2.00 0.263 3.80
When heated, colorless dinitrogen tetraoxide, N2O4(g), decomposes into red-brown nitrogen dioxide, NO2(g), according to the following...
When heated, colorless dinitrogen tetraoxide, N2O4(g), decomposes into red-brown nitrogen dioxide, NO2(g), according to the following reaction: N204 (g) + 2 NO2 (g) Suppose that 2.00 mol of N204(g) was placed into an empty 5.00-L flask and heated to 407 K. When equilibrium was attained, the concentration of red-brown NO2(g) was found to be 0.525 M. Calculate the equilibrium constant, K, for this reaction at 407 K. 3.80 2.00 0.500 0.263 0.525
Dinitrogen tetraoxide and nitrogen dioxide are two gases that exist in equilibrium at a range of temperatures. NO2 is a reddish brown gas while N204 is colorless. At High Temperature the red color is strong. || At Low Temperature the gas has less color. If we represent the equilibrium as: 2 NO2(g) = N204(9) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the...
Dinitrogen tetraoxide, a colorless gas, exists in equilibrium with nitrogen dioxide, a reddish brown gas. One way to represent this equilibrium is: N204(0) 2NO2(g) Indicate whether each of the following statements is true, T, or false, F. AT EQUILIBRIUM we can say that: _ 1. The concentration of NO2 is equal to the concentration of N204. _ 2. The rate of the dissociation of N2O4 is equal to the rate of formation of N204. 3. The rate constant for the...
please help
1. Dinitrogen tetraoxide (colorless gas) is converted to nitrogen dioxide (dark reddish brown gas) and is endothermic in the forward direction: A. Write a balanced equation. B. If a closed container containing a mixture of these two gases at equilibrium is a light brown in color when it is at room temperature (20.0°C), what change would you expect to observe if the container is placed in boiling water (100.0 °C)? c. If a closed container containing a mixture...
The second question is either true or false.
Dinitrogen tetraoxide is a colorless gas that dissociates into nitrogen dioxide, a reddish brown gas. N204(g) 2NO2(g) At equilibrium: 1. The forward reaction, N204g)2NO (g) A. Continues to happen B. Stops. C. Cannot tell without additional information. 2. Change in the concentration of N204 A. Continues to happen. B. Stops. C. Cannot tell without additional information. n tetraoxide, colorless gas, exists in equilibrium with nitrogen dioxide, a reddish brown gas. a One...
Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which is a reddish brown gas. N204(0) 22 NO2(g) Three experiments were run starting with different initial amounts of N204(9) ([N204]o in the table). The systems were allowed to reach equilibrium and the concentrations for each gas were measured (in units of moles/Liter). For each of the boxes below, calculate the ratios indicated by the column heading. This will show the mathematical relationship that exists...
Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which is a reddish brown gas. N204(9) 2 NO2(g) Three experiments were run starting with different initial amounts of N204(g) ([N2040 in the table). The systems were allowed to reach equilibrium and the concentrations for each gas were measured (in units of moles/Liter). For each of the boxes below, calculate the ratios indicated by the column heading. This will show the mathematical relationship that exists...
Dinitrogen tetraoxide is a colorless gas at room temperature. It can dissociate into nitrogen dioxide, which is a reddish brown gas. N2。4(g) 2 NO2(g) Three experiments were run starting with different initial amounts of N204(g) (IN204]o in the table). The systems were allowed to reach equilibrium and the concentrations for each gas were measured (in units of moles/Liter). For each of the boxes below, calculate the ratios indicated by the column heading. This will show the mathematical relationship that exists...
At High Temperature the red color is strong. At Low Temperature the gas has less color. If we represent the equilibrium as: N204() 2 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is decreased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature...
When heated, colorless dinitrogen tetraoxide, N204(8), decomposes into red-brown nitrogen dioxide, NO2(g), according to the following reaction: N204(g) 42 NO2 (g) Suppose that 2.00 mol of N204(8) was placed into an empty 5.00-L flask and heated to 407 K. When equilibrium was attained, the concentration of red-brown NO2(g) was found to be 0.525 M. Calculate the equilibrium constant, K., for this reaction at 407 K. 0.500 O 0.525 2.00 0.263 3.80
When heated, colorless dinitrogen tetraoxide, N2O4(g), decomposes into red-brown nitrogen dioxide, NO2(g), according to the following reaction: N204 (g) + 2 NO2 (g) Suppose that 2.00 mol of N204(g) was placed into an empty 5.00-L flask and heated to 407 K. When equilibrium was attained, the concentration of red-brown NO2(g) was found to be 0.525 M. Calculate the equilibrium constant, K, for this reaction at 407 K. 3.80 2.00 0.500 0.263 0.525
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