1 - Determine the enthalpy change for the reaction 2C(s) + 2H 2 O(g) ? CH 4 (g)
ID: 796898 • Letter: 1
Question
1 - Determine the enthalpy change for the reaction
2C(s) + 2H2O(g) ? CH4(g) + CO2(g)
Using the following enthalpy data
C(s) + H2O(g) ? CO(g) + H2(g) ?H = 131.3 kJ/mol
CO(g) + H2O(g) ? CO2(g) + H2(g) ?H = -41.2 kJ/mol
CH4(g) + H2O(g) ? 3H2(g) + CO(g) ?H = 206.1 kJ/mol
2 - Compute the enthalpy change for the reaction below using the heats of formation provided
Pb(NO3)2(aq) + 2NaI(aq) ? PbI2(s) + 2NaNO3(aq)
Explanation / Answer
1) The final reaction equation is
2 C(s) + 2H2O (g) --> CH4(g) + CO2(g)
You need to have the C(s) on the reactants side, and since there is only one reaction that has C(s), it must be reversed.
C(s) + H2O (g) --> CO(g) + H2 (g) change in H= 131 kJ
Since the reaction was reversed, the H has to change signs
However, there are two C(s) in the final reaction, so we multiply the reaction, and its H, by 2.
2C(s) + 2H2O (g) --> 2CO(g) + 2H2 (g) change in H= 262 kJ
Now we just get rid of all the compounds that aren't in the final reaction, and find out how to get CH4 (g) and CO2(g) to be in the products.
Put the reactions under each other to see:
2C(s) + 2H2O (g) --> 2CO(g) + 2H2 (g) change in H= 262 kJ
CO(g) + H2O(g) -> CO2(g) + H2(g) change in H= -41kJ
CO(g) + 3H2(g) -> CH4(g) + H2O(g) change in H = -206kJ
The CO(g) cancel each other out from products and reactants, the leftover H2O (g) cancels out, and the H2(g) cancels out so you're left with
2C(s) + 2H2O(g) -> CH4(g) + CO2(g)
But you need the change in H, so you add up all the changes in H from the reactions to get the final change in H.
262 kJ + (-41 kJ) + (-206 kJ) = 15 kJ
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