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