Calculate the delta H rxn for the following reaction:
CH4(g)+4Cl2(g)-->CCl4(g)+4HCl
Use the following reactions and given delta H's:
1) C(s)+2H2(g)-->CH4(g)
delta H= -74.6 kJ
2) C(s)+2Cl2(g)-->CCl4(g)
delta H= -95.7 kJ
3) H2(g)+Cl2(g)-->2HCl(g)
delta H= -184.6 kJ
Express your answer using one decimal place.
Well, calculating the delta H rxn involves using the given reactions and their delta H values to find the overall delta H of the given reaction. Let's break it down step by step.
For reaction 1), to make it match the desired reaction, we need to flip the equation and reverse the delta H sign:
C(s) + 2H2(g) <-- CH4(g)
delta H = 74.6 kJ (flipped sign)
For reaction 2), we don't need any changes:
C(s) + 2Cl2(g) --> CCl4(g)
For reaction 3), we also need to flip the equation and reverse the delta H sign:
H2(g) + Cl2(g) <-- 2HCl(g)
delta H = 184.6 kJ (flipped sign)
Now, we sum up the reactions to get the overall reaction:
C(s) + 2H2(g) + C(s) + 2Cl2(g) + 4H2(g) + Cl2(g) --> CH4(g) + CCl4(g) + 4HCl(g)
Now, let's add up the delta H values:
delta H rxn = (74.6 kJ) + (-95.7 kJ) + (4 * -184.6 kJ) = -1051.0 kJ
So, the delta H rxn for the given reaction is -1051.0 kJ. But hey, don't stress about it! Just remember, chemistry can be quite "reactive" sometimes!
To calculate the delta H_rxn for the given reaction, we will use the Hess's law. We can add up the given reactions and their respective delta H values to obtain the delta H_rxn for the desired reaction.
1) Multiply the first reaction by 4 to match the coefficients of HCl in the desired reaction.
4 * (C(s) + 2H2(g) --> CH4(g))
delta H_1 = 4 * (-74.6 kJ) = -298.4 kJ
2) Multiply the second reaction by 2 to match the coefficient of CCl4 in the desired reaction.
2 * (C(s) + 2Cl2(g) --> CCl4(g))
delta H_2 = 2 * (-95.7 kJ) = -191.4 kJ
3) Leave the third reaction as is since the coefficients match those in the desired reaction.
delta H_3 = -184.6 kJ
Now, add up the delta H values for each reaction.
delta H_rxn = (delta H_1) + (delta H_2) + (delta H_3)
delta H_rxn = (-298.4 kJ) + (-191.4 kJ) + (-184.6 kJ)
delta H_rxn = -674.4 kJ
Therefore, the delta H_rxn for the given reaction is -674.4 kJ.
To calculate the delta H rxn for the given reaction, we can use the concept of Hess's Law. Hess's Law states that if a reaction can be expressed as the sum of multiple reactions, then the enthalpy change of the overall reaction is the sum of the enthalpy changes of the individual reactions.
So, let's break down the given reaction into the three given reactions:
1) CH4(g) = C(s) + 2H2(g) (reverse of reaction 1)
delta H1 = +74.6 kJ (reverse sign)
2) 4Cl2(g) = 4HCl(g) + 2C(s) (multiply reaction 3 by 4 and reaction 2 by -2 to get the same number of moles of C(s))
delta H2 = -4 * (-184.6 kJ) + 2 * (-95.7 kJ) = +738.4 kJ - 191.4 kJ = +547.0 kJ
3) C(s) + 2Cl2(g) = CCl4(g) (same as reaction 2, just without HCl)
delta H3 = -95.7 kJ
Now, we can sum up these reactions to get the desired reaction:
CH4(g) + 4Cl2(g) = CCl4(g) + 4HCl(g)
delta H rxn = delta H1 + delta H2 + delta H3
= +(-74.6 kJ) + (+547.0 kJ) + (-95.7 kJ)
= +376.7 kJ
Therefore, the delta H rxn for the given reaction is +376.7 kJ (to one decimal place).
We'll reverse the 1st reaction:
CH4 --> C + 2H2 ; ΔH = +74.6
We'll use the 2nd equation as is:
C + 2Cl2 --> CCl4 ; ΔH = -95.7
We'll multiply the 3rd equation by 2:
2H2 + 2Cl2 --> 4HCl ; ΔH = -369.2
Now if we add the chemical equations, we'll end up with
CH4(g)+4Cl2(g)-->CCl4(g)+4HCl
Then add all the ΔH's to solve for the ΔH of this reaction.
Hope this helps~ `u`