A metal salt with the formula MCl2 crystallizes from water to form a solid with the composition MCl2 • 6 H2O. The equilibrium vapor pressure of water above this solid at 298 K is 18.3 mmHg. What is the value of ΔG for the reaction MCl2 • 6 H2O(s) ⇌ MCl2(s) + 6 H2O( g) when the pressure of water vapor is 18.3 mmHg? When the pressure of water vapor is 760 mmHg?

Video transcript

Hey everyone today, we're being asked to calculate this value of delta G. Or specifically, value of standard delta G. When a hydrated salt, magnesium, sulfate, Penta hydrate becomes magnesium sulfate and water. Specifically, we're being asked to calculate the standard delta G. At two separate vapor pressures. One when the vapor pressure is 21.8 M. M. G. M. M H. G. And one where the vapor pressure of water is 760 Hg. So let's go ahead and take a look at that first condition when it is at 28, mm Mercury. So earlier in the question, it is stated that the equilibrium vapor pressure of water is indeed 21.8 mm H. G. So this means therefore that At the pressure of 21. mm HG, the delta G Is equal to zero. Because we are at equilibrium at this vapor pressure at equilibrium. When the vapor pressure is 760 mercury, however, it gets a little more complicated. So let's go ahead and take a look at this. When the pressure is mercury, we still need to compare this back to our vapor pressure at equilibrium. So we need to find out what is our equilibrium vapor pressure in a. T. M. However, because we need to utilize the formula delta Qi Standard delta G of the reaction is equal to delta G. Of the reaction minus R T. Natural log of Q. And there's a few values, we need to know delta G of the reaction is our equilibrium delta G. R. As a gas constant temperature in kelvin and Q. Is the reaction quotient. And it is calculated as the a product of the product or the partial pressures of the products to their power of their coefficients over the product of the reactant, partial pressers, reactant to the power of their coefficients. However, this only applies for gasses only gas, which means that in our example, in our balanced chemical equation There's only one gas. It is the five Moles of H 20, which means the reaction quotient is therefore going to be Q. Is simply going to be the partial pressure of hydrogen. Or sorry, the water H 20 to the power of five because it has a coefficient of five, we still need to convert the equilibrium pressure. 21.8 millimeter H G two A. T. M. This is pretty easy. We just take 21.8 millimeter H. G. And recalled at 1 80 M. Is equivalent to 760 hg mm H. G. Which gives us a value of 0.28684 A. T. M. Therefore Q is equal to 0. to the power of five, which is equal to one 94, 2 times 10 to the negative 8th atmospheres. So the reason that I wrote out our equation earlier, like this with standard delta G. On the, solving for standard delta G is because we're being asked to find delta G, when the vapor pressure of water is 760 millimeters mercury. And since 760 millimeters mercury is just 1 80 M. That means we are at S. T. P. For that value. This right here is S. T. P. At standard temperature and pressure. So therefore we need to find the standard delta G. If it is any other value, we would have used delta G of the reaction is equal to the standard delta G of the reaction. Plus our Thailand. Que anyways with this we can go ahead and start solving So delta G. And let's scroll down a bit running out of space delta G. The standard Delta G is equal to zero. The equilibrium delta G minus the Gas constant eight 314 Joules Per Mole Kelvin. Remember we have to multiply this in order to get killer joules per mole kelvin because delta G is calculated in killing jewels. So we just multiply that by the conversion factor That one killer jewel is equal to 10 to the power of three jewels Times 298 Kelvin. Multiplied by the natural log of 1.942 times 10 to the negative eighth A. T. M. And finally this means that the standard delta G of the reaction. Or the reaction. I'm sorry the Delta G. When the reaction has the vapor pressure of 760, Mercury Is there for 44.0 killer jewels per mole. So when the vapor pressure is 21.8 It has a delta G of zero because that is its equilibrium and At standard temperature and pressure, or at 760 Mercury, the Delta G is 44 killer jewels per mole. I hope this helps. And I look forward to seeing you all in the next one.