Question

The standard enthalpies of formation of \(H_2O(l)\) and \(CO_2(g)\) are respectively \(-286\ \text{kJ mol}^{-1}\) and \(-394\ \text{kJ mol}^{-1}\). If the standard heat of combustion of \(CH_4(g)\) is \(-891\ \text{kJ mol}^{-1}\), then the standard enthalpy of formation of \(CH_4(g)\) is

• A. \(-75\ \text{kJ mol}^{-1}\)
• B. \(+75\ \text{kJ mol}^{-1}\)
• C. \(-211\ \text{kJ mol}^{-1}\)
• D. \(+211\ \text{kJ mol}^{-1}\)
• E. \(-1571\ \text{kJ mol}^{-1}\)

Hint:

For combustion, always write the balanced reaction first and then apply: \[ \Delta H = \Delta H_f(\text{products})-\Delta H_f(\text{reactants}) \]

Answer 1

The Correct Option is A

Concept: For combustion reaction: \[ CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l) \] \[ \Delta H_c = \sum \Delta H_f(\text{products})-\sum \Delta H_f(\text{reactants}) \]

Step 1: Substitute given values.
Let the enthalpy of formation of methane be \(x\). \[ -891 = [(-394)+2(-286)]-[x+0] \]

Step 2: Simplify.
\[ -891 = [-394-572]-x \] \[ -891 = -966-x \]

Step 3: Calculate \(x\).
\[ x=-966+891=-75\ \text{kJ mol}^{-1} \]