Consider the reaction
$$2\text{H}_2\text{S(g)} + 3\text{O}_2\text{(g)} \to 2\text{H}_2\text{O(l)} + 2\text{SO}_2\text{(g)}$$
The magnitude of enthalpy change for the reaction in kJ mol$$^{-1}$$ is __________. (Nearest integer)
Given :
$$\Delta_f H^\ominus(\text{H}_2\text{S}) = -20.1$$ kJ mol$$^{-1}$$
$$\Delta_f H^\ominus(\text{H}_2\text{O}) = -286.0$$ kJ mol$$^{-1}$$
$$\Delta_f H^\ominus(\text{SO}_2) = -297.0$$ kJ mol$$^{-1}$$

Using Hess's law,

$$\Delta_r H^\ominus=\sum \Delta_f H^\ominus(\text{products})-\sum \Delta_f H^\ominus(\text{reactants}).$$

For the reaction

$$2H_2S(g)+3O_2(g)\rightarrow2H_2O(l)+2SO_2(g),$$

the given standard enthalpies of formation are

$$\Delta_f H^\ominus(H_2S)=-20.1\ \text{kJ mol}^{-1},$$

$$\Delta_f H^\ominus(H_2O)=-286.0\ \text{kJ mol}^{-1},$$

$$\Delta_f H^\ominus(SO_2)=-297.0\ \text{kJ mol}^{-1},$$

$$\Delta_f H^\ominus(O_2)=0\ \text{kJ mol}^{-1}.$$

Substituting these values,

$$\Delta_r H^\ominus=[2(-286.0)+2(-297.0)]-[2(-20.1)+3(0)].$$

Simplifying,

$$\Delta_r H^\ominus=[-572.0-594.0]-[-40.2],$$

$$\Delta_r H^\ominus=-1166.0+40.2,$$

$$\Delta_r H^\ominus=-1125.8\ \text{kJ mol}^{-1}.$$

Hence, the magnitude of the enthalpy change is

$$|\Delta_r H^\ominus|=1125.8\ \text{kJ mol}^{-1}\approx1126\ \text{kJ mol}^{-1}.$$

Therefore, the required value is

$$1126.$$