Chemistry- thermochemistry

Question

How much glucose (in g) must be metabolized to completely evaporate 1.00 g of water at 37 °C, given that ∆_cH_glucose = -2803.00 kJ mol^-1? In other words how much glucose is needed to evaporate water at 37 °C where the vapour is at 37 °C too.

For H₂O(l) at 100 °C ∆_vH = 40.67 kJ mol^-1; C_p(H₂O(l)) = 75.3 J mol^-1 K^-1; C_p(H₂O(g)) = 33.6 J mol^-1 K^-1.

Answer 1

So, let's first look at the steps that are required to evaporate 1.00 g of water at 37ºC.

step 1. raise the temp of 1.00 g water from 37º to 100º

step 2. convert 1.00 g water @ 100º from liquid to gas (phase change)

Since the constants are given per mol H₂O, we'll change 1.00 g water to moles:

1.00 g H₂O x 1 mol H₂O / 18.0 g = 0.0555 moles H₂O

step 1. q = mC∆T = (0.0555 mols)(75.3 J/molº)(100º)

q = 418 J

step 2. q = m∆H_vap (phase change; no change in temperature)

q = (0.0555 mols)(40.67 kJ/mol)

q = 2.26 kJ

Add the two values together to get total heat needed (note one is in J and other is in kJ)

2.26 kJ + 0.418 kJ = 2.678 kJ needed

Now, we turn our attention to the data for glucose. We now know that we need 2.678 kJ of heat. From glucose, we can get 2803 kJ of heat from each mole. So, we perform the following calculations:

2.678 kJ needed x 1 mol glucose / 2803 kJ = 0.000955 moles of glucose needed to supply this amount of heat

Converting this to grams, we have...

0.000955 mols glucose x 180 g / mol = 0.172 grams glucose needed