Consider the following problem:
Target: 2NOCl (g) ¡æ N2 (g) + O2 (g) + Cl2 (g) ¥ÄHrxn= ?
Steps:
1. ¨ö N2(g) + ¨ö O2(g) ¡æ NO(g) ¥ÄH= 90.3 kJ
2. NO(g) + ¨ö Cl2(g) ¡æ NOCl(g) ¥Ä H= -38.6 kJ
As you determine the ¥ÄHrxn value for the target equation above, what must be done to step 2?
Question options:
a. Step 2 requires no change
b. Step 2 should be only multiplied by 2
c. Step 2 should be reversed and multiplied by 1/2.
d. Step 2 should be reversed and multiplied by 2.
e. Step 2 should be only reversed
To determine the ΔHrxn value for the target equation, you need to manipulate the given equations in such a way that they cancel out or combine properly to form the target equation. In this case, the target equation is:
2NOCl (g) ⇌ N2 (g) + O2 (g) + Cl2 (g)
Looking at step 2:
NO(g) + ½Cl2(g) ⇌ NOCl(g) ΔH = -38.6 kJ
To convert step 2 to match the target equation, you need to reverse the reaction and multiply it by a factor so that the coefficients of Cl2 and NOCl match the target equation.
The correct option is c. Step 2 should be reversed and multiplied by 1/2.
Reversing the equation:
NOCl(g) ⇌ NO(g) + ½Cl2(g)
Now, we need to multiply the reverse equation by 1/2 to match the coefficients:
(1/2)NOCl(g) ⇌ (1/2)NO(g) + ¼Cl2(g)
Now that the equations match, we can sum them up with step 1 to achieve the target equation:
(1/2)NOCl(g) + N2(g) + O2(g) + (1/4)Cl2(g) ⇌ (1/2)NO(g) + ¼Cl2(g) + N2(g) + O2(g)
Canceling out the components that appear on both sides:
(1/2)NOCl(g) + (1/4)Cl2(g) ⇌ (1/2)NO(g)
Now, we have achieved the target equation, and the ΔHrxn for the target equation is the sum of the ΔHrxn values for the individual reactions.