ADVANCED STUDY ASSIGNMENT 0.500 g of Ca is added to 50.0 mL of 1.0 M HCL. Once r

Question

ADVANCED STUDY ASSIGNMENT 0.500 g of Ca is added to 50.0 mL of 1.0 M HCL. Once reaction is complete, the temperature of the solution has risen from 25.2 degree C to 71.4 degree C. The reaction proceeds by the following equation: Ca(s) + 2 H+(aq) rightarrow Ca 2+(aq) + H2(g) Calculate the heat gained by the solution q = m middot s middot Delta T, where s = 4.184 J/g middot degree C m = mass in grams s = specific heat in J/g middot degree C Delta T = final temp. - initial temp. 50.5 times 4.184 times 46.2 = 9762 J Heat lost by the reaction: q r x n = -q soln Calculate the heat of reaction:

Explanation / Answer

qsol = msol x csol x dTsol = 50 x 4.184 x (74.4 - 25.2) ~ 9665.04 J = 9.665 kJ

since energy can only be transfered, not created, between a system (the reaction) and surroundings (the solution), the change in heat (q) is equal and opposite between them:

qrxn = -qsol = -9.665 kJ

dHrxn = kJ/mol Ca = -9.665 / (.5 g Ca / 40.078 g Ca/mol) ~ -9.665/.0125 ~ -774.7 kJ/mol

As for question 2 not shown in the picture: you want the dHC3H8 , so you have to reverse the third reaction.

The other 2 reactions are utilized and manipulated to isolate C3H8 as the only product. This means multiplying reaction 1 by 3, and reaction 2 by 4. This should leave you with an overall reaction of:

3C(s) + 4H2(g) --> C3H8

Using this, use the given information for the dH of every elementary reactions:

3(-393.5) + 4(-285.8) + 2199.0 ~ -124.7 kJ/mol

I round everything, so check my math. However, that is how you approach the question using Hess's Law with given enthalpy values for elementary reactions.

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