1.
The reaction for which we have to find Kc is
For a general reaction
the expression of Kc is
Hence, for our reaction Kc should be
Now, from the given reactions with Kc value
Similarly for
Now looking at the expression of Kc for the required reaction
We have to find in terms of .
Hence, we will first write the equation(0) in terms of equation (1) and (2).
O3 is in the reactant side in equation (0) and in the product side in equation (1)
Hence, we must reverse the equation in equation (1)
Hence, the new Kc1 of the reverse reaction is
There is only one mole of O3 in equation (0) in the product side, hence we must divide equation (3) by 2.
Hence, the new Kc1' is
Now our required equation (0) has 1 mole of NO on reactant side.
Hence, we must divide equation (2) by 2 to get 1 mole NO
Hence, Kc for equation (5).
Now, adding equation (4) and (5) we get the equation (0).
Hence, the Kc of equation (0) will be the product of equation (4) and equation (5) as shown below
Which is exactly what it should be.
Hence,
Given that
The required Kc is
hence, the answer is B .
2.
The given equilibrium is
Given that
we have to find out Kp, which is related to Kc by the following equation
is the number of moles of gaseous products minus number of moles of gaseous reactant.
Hence,
Gas constant R =
Hence, putting everything in
hence, the correct option is B.0.98 atm
help me by showing the work please Given the following information, Kc 2.0 x 10-38 Kc...
EX2 · Question 10 Determine the value of Kc for the following reaction if the equilibrium concentrations are as follows: [HCl]eq = 0.13 M; [HI]eq = 5.6 x 10-16M; [Cl2]eq = 0.0019 M. 2 HCl(g) + I2(s) ⇌ 2 HI(g) + Cl2(g) · Question 11 Consider the following reaction at equilibrium. What effect will be on the equilibrium when the pressure of the system rises from 4 atm to 12 atm? N2(g) + 3H2(g)) ⇌ 2NH3(g) · Question 12 What is the overall order of...
sic Equilibrium Calculations 25 of 33 below the reaction. Rank the extent of each rea refers to how much reactant is converted to product at equilibrium. equilibrium constant for the following two reactions. The equilibrium concentration of each reactant and product is given in the box action at equilibrium by calculating the equilibrium constant. Note that the "extent of each reaction Reaction 2 2N2(g) +O2(g) Reaction 1 CO(g) + O2(g) 2CO2(g) 2NO2(g) [CO] = 1.1 M [02] = 1.1 M...
Given the following reactions, (1) 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) Kc = 4.8×10^−10 (2) 2 NO(g) + O2(g) ⇌ 2 NO2(g) Kc = 9.1× 10^4 calculate Kc for the reaction below. (3) 2 NO2(g) ⇌ N2(g) + 2 O2(g) Kc = ???
Need answer and explanation please! b. What is the rate law for the overall-reaction? -I %10 pts) The relatonship between Kp and Kc is given by K,-K,(m". a. Calculate Kc for PCIs(g) → PC13(g) + Cl2(g) given KP-229 at 425°C. b. The reaction of nitrogen and hydrogen occurs according to the reaction N2 (g)+ 3H2 (g) → 2NH3 (g). Determine Kp for this reaction given Ke-6 x 105 at 25 °C x0602 An- nnr -1 b. What is the rate...
Give me the answer quickly please thank you 8. Using this data, 2 NO(g) + Cl2(g) = 2 NOCI(9) 2 NO2(g) = 2 NO(g) + O2(g) calculate a value for Kc for the reaction, NOCI(9) + 1/2O2(g) = NO2(g) + Kc = 3.20 x 10-3 Kc = 15.5 Cl2(g) a. 2.06 x 10-4 b. 4.84 x 103 c. 0.223 d. 4.49 e. 20.2
3. Given Kc or Kp for the following reactions, what is the value of Kp or Ke? (a) 12 (g) + Cl2 (a) 22ICI (g): Kc = 2.0 x105 at 25°C (b) N204(g) + 2NO2(0); Kc = 0.90 at 120 °C (c) CaCO3(s) = Cao (s) + CO2 (ox Kp = 1.67 x 102 at 740 °C 4. A container contains an equilibrium mixture of H2 (g), 12(g), and Hl) at 721 K. The concentration of each substance present at...
4. For the following chemical equilibrium, Kp-4.6 x 10-14 at 25°C, find the value of Kc for this reaction at 25°C SHOW ALL WORK 2 Cl2(g) + 2 H20(g)4 HCI(g)+ 02(9) R 0.083145 bar L/mol K 8.2057 x 10-2 L atm/mol K 8.3145 J/mol K Answer
i need help with everything please!! WWLWUNA HOSILNILNIUSPRING 2020) Chapter #14 The following practice problems from Chang's text are not to be submitted. The numbers in brackets refer to problems appearing at the end of the chapter. Equilibrium Constant Expressions (14.8, 14.9) Calculating Equilibrium Constants (14.13 to 14.24: Do as many as you can) Multiple Equilibria (14.29, 14.31) Kinetics and Chemical Equilibrium (14.35, 14.36) Calc. of Equilibrium Conc. (14.39 - 14.48) Do as many as you can. LeChatlier's Principle (14.54,...
2. How is the equilibrium-constant expression (Kc) for the reaction: 2NO(a) = N2() + Ke=0.145; related to the following reaction? O2 (a) N2(a) + O2(a) = 2NO(a) K=.............. (b) 4NOQ = 2Nz () + 2O2(g) Kos......... (c) NO) 1/2 N2(0)+ 1/2O2(0) K3= +++ (d) 1/2 N2(a) + 1/2O2(a) = NO) Ke=.............. 3. Given Kc or ko for the following reactions, what is the value of Koor K? (a) l2(g) + Cl2(a) = 2ICIOX Kc = 2.0 x105 at 25°C (b)...
Calculate Kc for the reaction 2CO(g) + 2Cl2(g) --> 2COCl2(g) given the following information: COCl2(g) --> CO(g) + Cl2(g) Kc = 1.0 x 10-4