Find the value of Q from the following equati | vedclass

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(A) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.Adding equations $(ii)$ and $

Vedclass · Accepted Answer

$ -(x+y) \ kJ $

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(A) According to Hess's Law,the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps.Adding equations $(ii)$ and $(iii)$:$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$Summing these gives:$C_{(s)} + \frac{1}{2} O_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} + CO_{(g)} - CO_{(g)}$ $\longrightarrow CO_{(g)} + CO_{2_{(g)}} - CO_{(g)}$$C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$Therefore,$Q = (-x) + (-y) = -(x+y) \ kJ$.

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