Question

In: Chemistry

At 473 K, for the elementary reaction: 2NOCl(g)k1⇌k22NO(g)+Cl2(g) k1=7.8×10−2L/mol⋅s and k2=4.7×102L2/mol2⋅s A sample of NOCl is...

At 473 K, for the elementary reaction:

2NOCl(g)k1⇌k22NO(g)+Cl2(g)

k1=7.8×10−2L/mol⋅s and
k2=4.7×102L2/mol2⋅s


A sample of NOCl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is 0.24 mol/L.

Find the concentration of NO.

Find the concentration of Cl2

Solutions

Expert Solution

k = k1 / k2

   = 7.8 x 10^-2 / 4.7 x 10^2

   = 1.66 x 10^-4

2 NOCl (g)   --------------->   2 NO (g) + Cl2 (g)

   a                                         0               0

a - 2x                                      2x               x

At equilibrium :

NOCl = 0.24

a - 2x = 0.24

K = [NO]^2[Cl2] / [NOCl]^2

1.66 x 10^-4 = (2x)^2 * x / (0.24)^2

x = 0.0134

concentration of NO = 0.027 M

concentration of Cl2 = 0.013 M

queen_honey_blossom answered 2 years ago
Next > < Previous

Related Solutions

At 473 K, for the elementary reaction 2NOCl(g)k1⇌k−12NO(g)+Cl2(g) k1=7.8×10−2L/mols and k−1=4.7×102L2/mol2s A sample of NOCl is...
At 473 K, for the elementary reaction 2NOCl(g)k1⇌k−12NO(g)+Cl2(g) k1=7.8×10−2L/mols and k−1=4.7×102L2/mol2s A sample of NOCl is placed in a container and heated to 473 K. When the system comes to equilibrium, [NOCl] is found to be 0.45 mol/L . Part A Find the concentration of NO. Part B Find the concentration of Cl2.
The reaction 2NO(g) + Cl2(g) --> 2NOCl(g) obeys the rate law rate = k [NO]2 [Cl2]....
The reaction 2NO(g) + Cl2(g) --> 2NOCl(g) obeys the rate law rate = k [NO]2 [Cl2]. The following mechanism is proposed: NO (g) + Cl2 (g) --> NOCl2(g) NOCl2(g) + NO (g) -->2NOCl (g) a) What would the rate law be if the first step was rate determining? b) Based on the observed rate law, what can be concluded about the relative rates of the 2 reactions?
2 NOCl(g) <=> 2 NO(g) + Cl2(g), K= 1.6E-5. Calculate the concentrations when: a). 6.0 mol...
2 NOCl(g) <=> 2 NO(g) + Cl2(g), K= 1.6E-5. Calculate the concentrations when: a). 6.0 mol of NO and 2.0 mol of Cl2 in a 1.2-L flask b).  2.2 mol of NOCl, 2.2 mol of NO, and 1.1 mol of Cl2 in a 1.0-L flask c).  1.01 mol/L concentration of all three gases Please show work and don't just solve the top problem. I need all three and I don't know how to correctly solve either of them!!
  2 NO(g) + Cl2(g) ⇄ 2 NOCl(g)    Kp = 1.40 × 10^8 A reaction vessel initially...
  2 NO(g) + Cl2(g) ⇄ 2 NOCl(g)    Kp = 1.40 × 10^8 A reaction vessel initially contains 3.0 atm of NO and 2.0 atm of Cl2. What is the pressure of Cl2(g) when equilibrium is reached?
  2 NO(g) + Cl2(g) ⇄ 2 NOCl(g)    Kp = 1.40 × 10^8 A reaction vessel initially...
  2 NO(g) + Cl2(g) ⇄ 2 NOCl(g)    Kp = 1.40 × 10^8 A reaction vessel initially contains 3.0 atm of NO and 2.0 atm of Cl2. What is the pressure of Cl2(g) when equilibrium is reached?
At 35oC, K = 1.6 x 10-5 for the following reaction 2NOCl(g) <----> 2NO(g) +   ...
At 35oC, K = 1.6 x 10-5 for the following reaction 2NOCl(g) <----> 2NO(g) +    Cl2(g) Calculate the concentration of all species at equilibrium for each of the following original mixtures. 1) 3.0 moles of pure NOCl in a 2.0 L flask 2) 2.0 moles of NOCl and 2 moles of NO in a 3 L flask 3) 3.0 moles of NOCl 1 mole of Cl2 in a 2.0 L flask Show work please
Consider the following equilibrium:           2NOCl(g)  2NO(g) + Cl2(g)with K = 1.6 × 10–5. 1.00 mole of pure...
Consider the following equilibrium:           2NOCl(g)  2NO(g) + Cl2(g)with K = 1.6 × 10–5. 1.00 mole of pure NOCl and 1.50 mole of pure Cl2 are placed in a 1.00-L container. Calculate the equilibrium concentration of NO(g).
NH4NO3(s) <---> N2O (g) + 2H2O(g) K = 0.665 at 525K If a 2L reaction vessel...
NH4NO3(s) <---> N2O (g) + 2H2O(g) K = 0.665 at 525K If a 2L reaction vessel contains 0.65 moles of N2O (g), 1.2 moles of 2H2O (g) and 1.3 moles of NH4NO3 (s), which of the following statements are correct? a. The amount of N2O (g) will increase and K will not change b. The amount of N2O (g) will increase and K will get larger Please explain!!!!!!!!!!!!!!! If the volume of the container is decreased, where will the equilibrium...
The rate constant of the elementary reaction 2HI(g) H2(g) +I2(g) is k = 1.97 L mol-1...
The rate constant of the elementary reaction 2HI(g) H2(g) +I2(g) is k = 1.97 L mol-1 s-1 at 604°C, and the reaction has an activation energy of 206 kJ mol-1. (a) Compute the rate constant of the reaction at a temperature of 767°C. L mol-1 s-1 (b) At a temperature of 604°C, 177 s is required for half of the HI originally present to be consumed. How long will it take to consume half of the reactant if an identical...
Part A Consider the second-order reaction: 2HI(g)→H2(g)+I2(g) Rate law: k[H]^2 k= 6.4*10^-9 (mol*s) at 500 K...
Part A Consider the second-order reaction: 2HI(g)→H2(g)+I2(g) Rate law: k[H]^2 k= 6.4*10^-9 (mol*s) at 500 K Initial rate = 1.6 * 10^-7 mol (l*s) What will be the concentration of HI after t = 3.65×1010 s ([HI]t) for a reaction starting under this condition? Part B In a study of the decomposition of the compound X via the reaction X(g)⇌Y(g)+Z(g) the following concentration-time data were collected: Time (min) [X](M) 0 0.467 1 0.267 2 0.187 3 0.144 4 0.117 5...
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT