From the following data at 25°C,

H2(g) + Cl2(g) → 2HCl(g) ΔH0 = −185 kJ
2H2(g) + O2(g) → 2H2O(g) ΔH0 = −483.7 kJ

Calculate ΔH0 at 25°C for the reaction below.
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)

Sorry ans hobe (185*2)-483.7=-113.7 ans,ok

Multiply equation 1 by 2 and reverse it. Add to equation 2. That will give you the equation you want but you need to go through the steps to confirm that.

When you multiply an equation, do the same to delta H. When you reverse and equation change the sign of delta H.

Hai,1no equestion *2, then, 1no-2no .

(185-483.7)=-298.7 ans.ok

To calculate ΔH° for the given reaction, we can use the concept of Hess's Law. According to Hess's Law, the enthalpy change for a reaction can be calculated by adding or subtracting the enthalpy changes of other reactions.

In this case, we can break down the reaction into two steps:
Step 1: Formation of 2HCl from H2 and Cl2
H2(g) + Cl2(g) → 2HCl(g) ΔH° = -185 kJ

Step 2: Formation of 2Cl2 and 2H2O from 4HCl and O2
2H2(g) + O2(g) → 2H2O(g) ΔH° = -483.7 kJ

Now, we need to reverse the first equation and multiply it by 2 to get the formation of Cl2 from 2HCl:
2HCl(g) → H2(g) + Cl2(g) ΔH° = +185 kJ

Finally, we can add these two equations to get the desired reaction:
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)

Adding the ΔH° values of the two equations together, we get:
(+185 kJ) + (-483.7 kJ) = -298.7 kJ

Therefore, the value of ΔH° for the given reaction is -298.7 kJ at 25°C.