With the help of thermochemical equations given below, determine `Delta_(r )H^(Θ)` at `298 K` for the following reaction:
`C("graphite")+2H_(2)(g) rarr CH_(4)(g),Delta_(r )H^(Θ) = ?`
`C("graphite")+O_(2)(g) rarr CH_(2)(g),
Delta_(r )H^(Θ) = -393.5 kJ mol^(-1)` ...(1)
`H_(2)(g) +1//2O_(2)(g) rarr H_(2)O(l)`,
`Delta_(r )H^(Θ) = -285.8 kJ mol^(-1)` ...(2)
`CO_2(2)(g)+2H_(2)O(l) rarr CH_(4)(g)+2O_(2)(g)`,
`Delta_(r )H^(Θ) = +890.3 kJ mol^(-1)` ...(3)

Explanation

To determine the standard enthalpy change (ΔrH°) for the reaction:

\[ C(\text{graphite}) + 2H_2(g) \rightarrow CH_4(g) \]

we will utilize the given thermochemical equations. Let's denote the enthalpy changes for the equations as follows:

1. \( C(\text{graphite}) + O_2(g) \rightarrow CH_2(g) \)
\[ \Delta H_1 = -393.5 \, \text{kJ/mol} \]
...