6 kJ/mol$
If we divide the equation by 2, we get:
$C_4H_{10}(g) + \frac{13}{2} O_2(g) \rightarrow 5 H_2O(l) + 4 CO_2(g)$
Since $\Delta H$ is an extensive property, when the reaction is divided by 2, the value of $\Delta H$ is also divided by 2.
Therefore, the new
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