The initial concentration of hi is 0100 m and kc is 011 for...

The initial concentration of HI is 0.100 M and Kc is 0.11 for the reaction: 2HI(g) ⇌ H2(g) + I2(g) What is the equilibrium concentration of HI? The value of Kc at 1100 K for the following chemical equilibrium is 7.14 x 10^9: 2NO(g) + O2(g) ⇌ 2NO2(g) What is the value of Kc at 1100 K for the following reaction? NO2(g) ⇌ NO(g) + 1/2O2(g) Consider the following equilibrium: CO(g) + 3 H2(g) ⇌ CH4(g) + H2O(g) If Kp = 1.61 × 10^-5 at 1400.0 K, calculate Kc. The decomposition of NOCl to form NO and Cl2, shown below: 2NOCl(g) ⇌ 2NO(g) + Cl2(g) has a Kp value of 1.6 × 10^-5 at some temperature. If 0.500 atm of NOCl is placed in a closed vessel and allowed to come to equilibrium, what is its approximate equilibrium partial pressure?

Transcript

00:01 Hello students, this question consists of four parts.

00:03 In the first part, the equation for the dissociation of hydrogen iodide has been given and its dissociation constant and the initial concentration has been given.

00:11 We have to find out its equilibrium concentration.

00:15 So, at first, we are going to derive the kc expression for this which is given by the product of concentration of the product divided by the product of concentration of the reactant with the number of moles raised to the power.

00:30 Substitute in the equilibrium concentration.

00:33 The equilibrium concentration of the h2 and i2 will be equal to x.

00:37 X into x it will become x square divided by 0 .1 minus x.

00:42 This will be equal to 0 .11 the whole square.

00:47 So, minus x is negligible because it is very much less than 0 .1.

00:53 Therefore, this equation will become 0 .11 will be equal to x square divided by 0 .1 v square.

01:01 On solving this, we will get the x value will be equal to 0 .033.

01:09 So, we know that equilibrium concentration of the hi will be equal to 0 .1 minus x.

01:16 Therefore, 0 .1 minus 0 .033.

01:20 On solving this, we will get the equilibrium concentration of the hi will be equal to 0 .067 m.

01:31 This is our answer.

01:33 Coming to the second part of the question.

01:35 So, the dissociation constant for the equation has been given.

01:39 We have to find out the kc dash of this equation.

01:42 Both are taken place at the same temperature.

01:45 So, at first, we are going to try to derive the second equation from the first equation.

01:50 So, let this equation be a.

01:52 We are going to divide it by 2 so that we will get no plus half o2 gives us no2.

02:00 Now, we are going to reverse this equation.

02:02 Therefore, minus a by 2 we will get no2 gives no plus half o2.

02:12 So, we have achieved the targeted equation which is obtained by reversing the first equation and dividing by 2.

02:19 Therefore, kc dash is obtained by reversing the kc value divided by 2...

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