Consider the system ag bg answer the following questions a...

Consider the system: A(g) B(g) Answer the following questions (a) to (c): (a) At 25°C, assume that ΔG°A = 7358 J/mol and ΔG°B = 12402 J/mol. Calculate the value of the equilibrium constant for this reaction. (b) A non-equilibrium mixture of 1.00 mol of A(g) (partial pressure 1.00 atm) and 1.00 mol of B(g) (partial pressure 1.00 atm) is allowed to equilibrate at 25°C. Calculate the partial pressure of A(g) at equilibrium. (c) Calculate the partial pressure of B(g) at equilibrium.

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00:01 Okay, consider system you have a becomes b.

00:08 So the first question is to calculate the equation constant.

00:13 So first we need to calculate the delta g of this reaction.

00:19 So the delta g of this reaction equals to the delta g b minus delta g.

00:27 Or gb of minus ga, more simply.

00:35 So that's equal to 1, 2.

00:39 2402 minus 3 7358 so that's equal to about 5 044 droop per more so we know that the delta g equals to e2 uh sorry the quorum constant k equals to e2 minus delta g over rt.

01:26 So here just plug in, delta g is minus 5 over 44, r is 8 .314, t is 298 .15.

01:45 So that's equal to about 0 .13, the k.

02:00 And the b, a non -equipimate mix feature of the a and b, is allowed to equilibrium at 25.

02:13 So again we can write down the equation and draw an ice table.

02:19 Initially you have one more the a, one more the b, and the change right definitely will go to change to the equivalent.

02:28 So suppose that a was consumed, b was produced, so because minus x plus x.

02:36 So in the end you have 1 minus x 1 plus x as b.

02:42 And in quid constant k equals to 1 plus x over 1 minus x.

02:49 And that's equal to this 0 .13.

02:56 So 1 plus x equals to 0 .13 minus 0 .13 x...

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