Consider the following thermochemical equations.
Reaction ΔrH° / kJ mol−1
HBr(g) ⟶ H(g) + Br(g) 365.7
H2(g) ⟶ 2 H(g) 436.0
Br2(g) ⟶ 2 Br(g) 193.9
What is ΔrH° for the reaction below, in kJ mol-1?
HBr(g) ⟶ ½ H2(g) + ½Br2(g)
what is 365.7 -1/2 (436.0) - 1/2 (193.9)
Well, it seems like we have quite the chemistry situation here. We need to find the ΔrH° for the reaction:
HBr(g) ⟶ ½ H2(g) + ½Br2(g).
Let's break it down step by step.
The ΔrH° for the given reactions are:
HBr(g) ⟶ H(g) + Br(g) : 365.7 kJ mol-1
H2(g) ⟶ 2 H(g) : 436.0 kJ mol-1
Br2(g) ⟶ 2 Br(g) : 193.9 kJ mol-1
Now, we can use these values to calculate the ΔrH° for the desired reaction.
First, let's adjust the coefficients so that we have equal numbers of hydrogen and bromine atoms on both sides of the equation.
Starting with HBr(g) ⟶ ½ H2(g) + ½Br2(g), we can multiply the equation by 2 to eliminate the ½ coefficients:
2 HBr(g) ⟶ H2(g) + Br2(g).
Now, let's add the ΔrH° of the individual reactions to find the ΔrH° for the desired reaction:
ΔrH° = ΔrH° (H2) + ΔrH° (Br2) - ΔrH° (HBr)
= (436.0) + (193.9) - (365.7)
= 264.2 kJ mol-1.
So, the ΔrH° for the reaction HBr(g) ⟶ ½ H2(g) + ½Br2(g) is 264.2 kJ mol-1.
I hope that helps! Chemistry can be a real balancing act... quite literally.
To determine the ΔrH° for the reaction:
HBr(g) ⟶ ½ H2(g) + ½Br2(g)
We can use the thermochemical equations given to calculate the desired ΔrH°.
First, we can recognize that the given equation:
H2(g) ⟶ 2 H(g)
Is the reverse of the desired equation. Therefore, we can reverse the sign of the ΔrH° for that equation:
ΔrH° = -436.0 kJ mol-1
Next, we observe that the equation:
HBr(g) ⟶ H(g) + Br(g)
is the same as half of the desired equation. Therefore, we can divide the ΔrH° for that equation by 2:
ΔrH° = 365.7 kJ mol-1 / 2 = 182.85 kJ mol-1
Lastly, we note that the equation:
Br2(g) ⟶ 2 Br(g)
is also the reverse of half of the desired equation. So, we need to reverse the sign of the ΔrH° for that equation, as well as divide it by 2:
ΔrH° = -(193.9 kJ mol-1) / 2 = -96.95 kJ mol-1
Now, we can sum up the ΔrH° values for the individual reactions to get the ΔrH° for the desired reaction:
ΔrH° = (-436.0 kJ mol-1) + (182.85 kJ mol-1) + (-96.95 kJ mol-1)
ΔrH° = -349.1 kJ mol-1
Therefore, ΔrH° for the reaction:
HBr(g) ⟶ ½ H2(g) + ½Br2(g)
is -349.1 kJ mol-1.
To find the ΔrH° for the reaction HBr(g) ⟶ ½ H2(g) + ½Br2(g), you can use Hess's Law. Hess's Law states that the overall enthalpy change of a reaction is independent of the reaction pathway and is equal to the sum of the enthalpy changes of the individual reactions.
In this case, we can find the ΔrH° for the target reaction by combining the given thermochemical equations.
Let's break down the target reaction into two steps:
Step 1: HBr(g) ⟶ H(g) + Br(g) (given ΔrH° = 365.7 kJ mol−1)
Step 2: H(g) + Br(g) ⟶ ½ H2(g) + ½Br2(g)
Since the target reaction is the combination of these two steps, the overall enthalpy change for the target reaction is the sum of the enthalpy changes for each step.
ΔrH°(target) = ΔrH°(step 1) + ΔrH°(step 2)
Substituting the given values:
ΔrH°(target) = 365.7 kJ mol−1 + ΔrH°(step 2)
To find ΔrH°(step 2), we need to manipulate the given equations to match the desired reaction.
From the given equations:
H2(g) ⟶ 2 H(g) (ΔrH° = 436.0 kJ mol−1)
Br2(g) ⟶ 2 Br(g) (ΔrH° = 193.9 kJ mol−1)
We can multiply both sides of the second equation by ½ to get:
½ Br2(g) ⟶ Br(g)
Now we can add the two equations together:
H2(g) + ½ Br2(g) ⟶ 2 H(g) + Br(g)
Since we want to form ½ H2(g) and ½ Br2(g), we can divide the equation by 2:
½ H2(g) + ½ Br2(g) ⟶ H(g) + ½ Br(g)
The enthalpy change for this new equation is the negative of the sum of the enthalpy changes of the original equations:
ΔrH°(step 2) = -[(436.0 kJ mol−1) + (193.9 kJ mol−1)]
Now substitute this value into the equation for ΔrH°(target):
ΔrH°(target) = 365.7 kJ mol−1 + [-[(436.0 kJ mol−1) + (193.9 kJ mol−1)]]
Simplifying the equation:
ΔrH°(target) = 365.7 kJ mol−1 - 436.0 kJ mol−1 - 193.9 kJ mol−1
ΔrH°(target) = -264.2 kJ mol−1
Therefore, the ΔrH° for the reaction HBr(g) ⟶ ½ H2(g) + ½Br2(g) is -264.2 kJ mol−1.