A mixture of 0100 mol of no 0050 mol of h2 and 0100 mol of...

A mixture of 0.100 mol of NO, 0.050 mol of H2, and 0.100 mol of H2O is placed in a 0.50 L vessel. The following equilibrium is established: 2 NO (g) + 2 H2 (g) ⇌ N2 (g) + 2 H2O (g) At equilibrium [NO] = 0.130 M. Calculate the equilibrium concentrations of: H2 N2 a) The following reaction is at equilibrium. CH4 (g) + H2O (l) ⇌ CO (g) + 3 H2 (g) ΔH = +203 kJ Use Increase, Decrease or No Change to describe what will happen when: -moles of CO when H2 is added -Kc when the Temperature increases

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00:01 Dear students, in a vessel of 0 .5 litre, 100 mol of no, 0 .05 mole of hydrogen and 0 .10 mole of hydrogen are introduced and they reacted, the no and h2 reacted and come to equilibrium as given in the equation.

00:18 At equilibrium it is found, no molar concentration is 0 .13.

00:22 Then what is the molar concentration of hydrogen and nitrogen at equilibrium? this is our question.

00:27 Now initial concentration let us find out initial concentration of no in the beginning.

00:35 In the beginning it is, initial concentration is we have 100 moles in 0 .5 liter.

00:42 So, moles per liter will be 100 by 0 .5 multiplied by 1 equal to 200 mole per liter.

00:53 Then initial concentration of hydrogen is 0 .05 divided by 0 .5 liter multiplied by 1 liter.

01:05 So it is coming 0 .1 mole per liter.

01:13 Now initial concentration of water, initial concentration of water molar concentration equal to it is 0 .1 in 0 .5 liter.

01:27 So 0 .1 .1.

01:28 1 .5 multiplied by 1 equal to 1 by 5 equal to 0 .2 mole per liter.

01:40 It is actually 0 .1, 0 .1.

01:44 So if it is 0 .1, it will be how much? it will be 0 .2.

01:52 So 0 .2 mole per liter, initial concentration of nau.

01:55 0 .1 is hydrogen and 0 .2 is water.

01:59 Now, at equilibrium 0 .13 mole per liter is there...

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