At 298 K, the enthalpy change (in kJ) for th | vedclass

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(B) To find the enthalpy change for the reaction $CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$, we manipulate the given equations:Equation

Vedclass · Accepted Answer

$-496$

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(B) To find the enthalpy change for the reaction $CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$, we manipulate the given equations:Equation $(3): CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} ; \Delta H^{\ominus} = -890 \ kJ$Reverse Equation $(2): CO_{2(g)} \rightarrow C_{(s)} + O_{2(g)} ; \Delta H^{\ominus} = +394 \ kJ$Adding these two equations:$CH_{4(g)} + 2O_{2(g)} + CO_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} + C_{(s)} + O_{2(g)}$Simplifying, we get:$CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$The enthalpy change is the sum of the enthalpy changes of the steps:$\Delta H = -890 \ kJ + 394 \ kJ = -496 \ kJ$.

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