Homework Answers
The given equilibrium is
N2O4 (g) 
2NO2 (g)
The given equilibrium concentrations are as follows
[N2O4] = 0.800 M and [NO2] = 0.400 M
The equilibrium concentration can be written as
Keq = [NO2]2/[N2O4]
= (0.400)2/(0.800) = 0.2
Thus the answer is option B
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Dinitrogen tetroxide decomposes to produce nitrogen dioxide: N204 (9) - 2 NO2 (g) Calculate the equilibrium...
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
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2. For the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N204 = 2 NO2 Suppose that both the forward and reverse reactions are elementary processes with rate constants of k, and kr respectively, and the equilibrium constant for the process is 14.48 at 298 K. If the rate of the reverse reaction is 1324 M/s when (NO2) = 0.60 M, what is the half-life for the decomposition of dinitrogen tetroxide (the forward reaction) at the same temperature?
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
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Dinitrogen tetroxide decomposes to nitrogen dioxide. Write the balanced chemical equation for this reaction. Start with pure N2O4 at 1.00 atm. Find the equilibrium partial pressures of N2O4 and NO2. The equilibrium constant for this reaction is K = 11.
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The density of a 0.84 M aqueous sugar (C12H22O11) solution is 1.12 g/mL at 25°C. What...
The density of a 0.84 M aqueous sugar (C12H22O11) solution is 1.12 g/mL at 25°C. What is the molality? The molar mass of C12H22O11 = 342.3 g/mol. Dinitrogen tetroxide decomposes to produce nitrogen dioxide: N2O4 (g) ↔ 2 NO2 (g) Calculate the equilibrium constant for the reaction given the equilibrium concentrations at 100°C: [N2O4] = 0.800 M and [NO2] = 0.400 M
1) Dinitrogen tetroxide partially decomposes according to the following equilibrium : N204 (9) = 2'N02 (9)....
1) Dinitrogen tetroxide partially decomposes according to the following equilibrium : N204 (9) = 2'N02 (9). A 1.00 L-flask is charged with 0.0400 mol of N204. At equilibrium clt 373 K, 0.0055 mol of N204 remains. Key for this reaction is -
Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g)→2NO2(g)ΔHorxn= 55.3kJ At 298 K, a reaction vessel initially contains...
Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g)→2NO2(g)ΔHorxn= 55.3kJ At 298 K, a reaction vessel initially contains 0.100 atm of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2. What percentage of N2O4 decomposes at 380 K ? Assume that the initial pressure of N2O4 is the same (0.100 atm).
Question 7 At low temperature nitrogen dioxide molecules join together to form dinitrogen tetroxide. 2 NO2(g)...
Question 7 At low temperature nitrogen dioxide molecules join together to form dinitrogen tetroxide. 2 NO2(g) + N204(9) (low temperature) A sample of NO2 sealed inside a glass bulb at 23 °C gave a pressure of 673 Torr. Lowering the temperature to -5 °C converted the NO2 to N204. What was the final pressure (in Torr) inside the bulb? Torr
Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g)→2NO2(g) ΔrH∘=55.3kJmol−1 At 298 K, a reaction vessel initially contains...
Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g)→2NO2(g) ΔrH∘=55.3kJmol−1 At 298 K, a reaction vessel initially contains 0.100 bar of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2. What percentage of N2O4 decomposes at 369 K? Assume that the initial pressure of N2O4 is the same (0.100 bar)
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
2. For the equilibrium between dinitrogen tetroxide and nitrogen dioxide: N204 = 2 NO2 Suppose that both the forward and reverse reactions are elementary processes with rate constants of k, and kr respectively, and the equilibrium constant for the process is 14.48 at 298 K. If the rate of the reverse reaction is 1324 M/s when (NO2) = 0.60 M, what is the half-life for the decomposition of dinitrogen tetroxide (the forward reaction) at the same temperature?
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
1) Dinitrogen tetroxide partially decomposes according to the following equilibrium : N204 (9) = 2'N02 (9). A 1.00 L-flask is charged with 0.0400 mol of N204. At equilibrium clt 373 K, 0.0055 mol of N204 remains. Key for this reaction is -
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