A) \[-321.30-300R\]
B) \[-321.30+300R\]
C) \[-321.30-150R\]
D) \[-321.30+900\text{ }R\]
Correct Answer: C
Solution :
\[{{C}_{6}}{{H}_{5}}COOH(s)+\frac{15}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}\] \[7C{{O}_{2}}(g)+3{{H}_{2}}O(l)\] \[\Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}}=7-\frac{15}{2}=\frac{-1}{2}\] \[{{q}_{p}}={{q}_{v}}+\Delta {{n}_{g}}(RT)\] \[=-321.30+\left( -\frac{1}{2} \right)300R\] \[=-321.30-150\text{ }R\]You need to login to perform this action.
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