Question:
At \(298\,K\), the value of \((\Delta H-\Delta U)\) for the combustion of 1 mole of
\[
C_4H_{10}(g)
\]
is \(x\) kJ and for the combustion of 1 mole of glucose, the value of
\[
(\Delta H-\Delta U)
\]
is \(y\) kJ. The value of \((x-y)\) (in kJ) is
\[
(R=8.3\ \text{J K}^{-1}\text{mol}^{-1})
\]
At \(298\,K\), the value of \((\Delta H-\Delta U)\) for the combustion of 1 mole of
\[
C_4H_{10}(g)
\]
is \(x\) kJ and for the combustion of 1 mole of glucose, the value of
\[
(\Delta H-\Delta U)
\]
is \(y\) kJ. The value of \((x-y)\) (in kJ) is
\[
(R=8.3\ \text{J K}^{-1}\text{mol}^{-1})
\]
Show Hint
For chemical reactions involving gases:
\[
\boxed{\Delta H-\Delta U=\Delta n_gRT}
\]
Only gaseous species are counted while calculating \(\Delta n_g\). Solids and liquids are ignored.
Updated On: Jul 29, 2026
- \(+8.657\)
- \(-8.657\)
- \(-9.659\)
- \(+4.329\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
For reactions involving gases,
\[
\Delta H-\Delta U=\Delta n_gRT,
\]
where
\[
\Delta n_g
=
(\text{moles of gaseous products})
-
(\text{moles of gaseous reactants}).
\]
Step 1: Find \(x\) for combustion of butane. Balanced equation: \[ C_4H_{10}(g)+\frac{13}{2}O_2(g) \rightarrow 4CO_2(g)+5H_2O(l) \] Therefore, \[ \Delta n_g = 4-\left(1+\frac{13}{2}\right). \] \[ = 4-7.5 = -3.5. \] Hence, \[ x = \Delta n_gRT = (-3.5)(8.3)(298). \] \[ x=-8656.9\ \text{J} =-8.657\ \text{kJ}. \]
Step 2: Find \(y\) for combustion of glucose. Balanced equation: \[ C_6H_{12}O_6(s)+6O_2(g) \rightarrow 6CO_2(g)+6H_2O(l) \] Thus, \[ \Delta n_g = 6-6 = 0. \] Therefore, \[ y=0. \]
Step 3: Calculate \(x-y\). \[ x-y = (-8.657)-0. \] \[ x-y=-8.657\ \text{kJ}. \]
Final Answer: \[ \boxed{-8.657\ \text{kJ}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Find \(x\) for combustion of butane. Balanced equation: \[ C_4H_{10}(g)+\frac{13}{2}O_2(g) \rightarrow 4CO_2(g)+5H_2O(l) \] Therefore, \[ \Delta n_g = 4-\left(1+\frac{13}{2}\right). \] \[ = 4-7.5 = -3.5. \] Hence, \[ x = \Delta n_gRT = (-3.5)(8.3)(298). \] \[ x=-8656.9\ \text{J} =-8.657\ \text{kJ}. \]
Step 2: Find \(y\) for combustion of glucose. Balanced equation: \[ C_6H_{12}O_6(s)+6O_2(g) \rightarrow 6CO_2(g)+6H_2O(l) \] Thus, \[ \Delta n_g = 6-6 = 0. \] Therefore, \[ y=0. \]
Step 3: Calculate \(x-y\). \[ x-y = (-8.657)-0. \] \[ x-y=-8.657\ \text{kJ}. \]
Final Answer: \[ \boxed{-8.657\ \text{kJ}} \] \[ \boxed{\text{Answer = (B)}} \]
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