Part B The following reaction was performed in a sealed...

Part B The following reaction was performed in a sealed vessel at 757 °C: $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ Initially, only $H_2$ and $I_2$ were present at concentrations of $[H_2] = 3.00M$ and $[I_2] = 2.15M$. The equilibrium concentration of $I_2$ is 0.0400 M. What is the equilibrium constant, $K_c$, for the reaction at this temperature? Express your answer numerically. $K_c = $

Transcript

00:01 Hello everyone, we are provided with this reaction and we have to find the equilibrium constant for this reaction.

00:06 So what will be the equilibrium constant for this reaction that would be equal to h2 into i2 divided by hi to the power of whole to concentration.

00:20 We have the value of h2 and i2 but we don't have the value of hi.

00:25 So for this first we have to calculate the value of hi.

00:29 Now according to the question that is atq, we are provided with the reaction h2 plus i2 which in the equilibrium would give 2hi.

00:41 If initially and at the initial stage we say that we have x moles of h2 and x moles of i2 and 2x moles of ix.

00:52 Therefore, we can say that now this much moles we have at the initial concentration then the final concentration would be since the moles for x2 and i2 are already provided to us.

01:05 So what would be the final one that is 6 .05 minus of x and i would be 6 .05 minus of x and what would be 2 into hi.

01:18 So 2 into hi would remain the same since we don't have this.

01:22 So it would be basically the concentration if i talk then 2x square or 2x.

01:28 Now on the basis of this we can write the equilibrium constant for this reaction.

01:32 So the k for this reaction would be hi square divided by h2 into i2.

01:46 So we can put the values over here.

01:48 So it would be equal to 2x square divided by 0 .65 minus x into 0 .65 minus x.

02:03 So the value of the value of k for this reaction has already been provided to us which is equal to 0 .0049.

02:11 So this would be equal to 2x square divided by we can take the square of this since it is the multiplication of the same number.

02:20 So it is 0 .65 minus of x to the power of square.

02:28 So now we can solve the value of x.

02:30 On solving the value of x it comes out to be 0 .038 mole per liter...

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