What is the difference between Δ H and Δ U at 298 K for the following reaction? C 2 H 4(g)+3 O 2(g) arrow 2 CO 2(g)+2 H 2 O ( I ) Δ H=-1410.0 kJ

Question

What is the difference between Δ H and Δ U at 298 K for the following reaction? C 2 H 4(g)+3 O 2(g) arrow 2 CO 2(g)+2 H 2 O ( I ) Δ H=-1410.0 kJ (A) -14.8 k J (B) -2.45 k J (C) -4.95 k J (D) -7.30 kJ

Tardigrade Admin · Accepted Answer

In the given reaction, C 2 H 4(g)+3 O 2(g) arrow 2 CO 2(g)+2 H 2 O (l) Δ H=-1410.0 kJ text For ideal gas, Δ H=Δ U+R T(Δ ng) text where, R=8.314 J / text molK T=298 K Δ ng= text number of gaseous products - number of gaseous reactants =2-4=-2 ∴-1410=Δ U+[8.314 × 298(-2)] Δ U=-1405.045 kJ ∼ e q-1405.05 kJ text Now, Δ H-Δ U=-1410-(-1405.05)=-4.95 kJ

Q. What is the difference between $Δ H$ and $Δ U$ at $298 K$ for the following reaction?
$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (I)$
$Δ H = - 1410.0 k J$

$- 14.8 k J$

$- 2.45 k J$

$- 4.95 k J$

$- 7.30 k J$

Q. What is the difference between ΔH and ΔU at 298K for the following reaction? C2​H4​(g)+3O2​(g)→2CO2​(g)+2H2​O(I) ΔH=−1410.0kJ

−14.8kJ

−2.45kJ

−4.95kJ

−7.30kJ

Q. What is the difference between $Δ H$ and $Δ U$ at $298 K$ for the following reaction?
$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (I)$
$Δ H = - 1410.0 k J$

$- 14.8 k J$

$- 2.45 k J$

$- 4.95 k J$

$- 7.30 k J$