When 0.250 mol of CH4(g) reacts with excess Cl2(g) at constant pressure according to the chemical equation shown below, 177 kJ of heat are released. Calculate the value of Delta H for this reaction, as written.
2CH4(g)+3Cl2(g)-->2CHCl3(l)+3H2(g)
delta H=?
moles = grams/molar mass
0.25 = g/16 and grams = 0.25 x 16 = 4 grams.
The equation as written is 2 moles CH4 so that is 16 x 2 = 32 grams.
177 kJ (for the reaction for the 4 grams) x (32/4) = ??
I get 1416 kJ. If you rounded your answer because of significant figures, showing it as 1420 won't do it. Most texts recommend writing it as 1.42 x 10^3 kJ.
hhmm would it be 4x that amount and not 8? im confused
i guess the teacher put a bad answer bank none of the answer are 1416 but i get what you are saying i will bring this up to him thanx
Try 1.42 x 10^3 kJ and let me know.
I am hoping that this could help you. I had to shorten the link because it was reaching the character limit.
bit.ly/3dD2rEp
To calculate the value of ΔH for the given reaction, we need to use the concept of stoichiometry and apply the following equation:
ΔH = q / (moles of limiting reactant)
Given the information:
- The reaction releases 177 kJ of heat.
- The balanced chemical equation is 2CH4(g) + 3Cl2(g) -> 2CHCl3(l) + 3H2(g).
We need to determine the moles of the limiting reactant in order to find ΔH.
Step 1: Find the moles of CH4:
Since we are given 0.250 mol of CH4, we can say that there are also 0.250 mol of CH4 reacting in this equation.
Step 2: Use stoichiometry to find the moles of Cl2 consumed:
From the balanced equation, we can see that the ratio between CH4 and Cl2 is 2:3.
So, for every 2 moles of CH4 reacting, 3 moles of Cl2 react.
Thus, for 0.250 mol of CH4 consumed, we can calculate the moles of Cl2 consumed using the ratio:
(0.250 mol CH4) x (3 mol Cl2 / 2 mol CH4) = 0.375 mol Cl2.
Since Cl2 is in excess, it is not the limiting reactant, and we can consider the moles of CH4 as the limiting reactant.
Step 3: Calculate ΔH:
Using the formula mentioned earlier, we divide the heat released (177 kJ) by the moles of the limiting reactant (0.250 mol):
ΔH = 177 kJ / 0.250 mol = -708 kJ/mol.
Therefore, the value of ΔH for this reaction is -708 kJ/mol, as written in the balanced chemical equation.