Given- C2H2g- C2H4g- H-175 kJ mol-1- H-fC2H4-g-50kJ mol-1- - H-fH2O-l-280 kJ mol-1- - H-fCO2g-390 kJ mol-1If - H- is enthalpy of combustion in kJ mole-1 of C2H2g- then calculate the value of - - H-257 -

C_2H_2(g)+H_2(g)→ C_2H_4(g) Given, nbsp;Δ H^∘_f(C_2H_4,g)=50 kj mol^ 1 ∴ 175= (Δ_fH)_C_2H_2+50 ⇒ (Δ_fH)C_2H_2=225 kj mol^ 1 For enthalpy of combustion the cor ...

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Byju's Answer

Standard XII

Chemistry

Limitations of First Law

Given, C2H2g→...

Question

Given, $C_{2} H_{2} (g) \to C_{2} H_{4} (g) : Δ H^{\circ} = - 175 {kJ mol}^{- 1}$
$Δ H_{f (C_{2} H_{4}, g)}^{\circ} = 50 kJ mo l^{- 1}; Δ H_{f (H_{2} O, l)}^{\circ} = - 280 {kJ mol}^{- 1}; Δ H_{f (C O_{2} g)}^{\circ} = - 390 {kJ mol}^{- 1}$
If $Δ H^{\circ}$ is enthalpy of combustion (in ${kJ mole}^{- 1}$ of $C_{2} H_{2} (g)$ , then calculate the value of $∣ ∣ ∣ \frac{Δ H^{\circ}}{257} ∣ ∣ ∣$

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Solution

$C_{2} H_{2} (g) + H_{2} (g) \to C_{2} H_{4} (g)$
Given, $Δ H_{f (C_{2} H_{4}, g)}^{\circ} = 50 {kj mol}^{- 1}$
$∴ - 175 = - (Δ_{f} H)_{C_{2} H_{2}} + 50$
$\Rightarrow (Δ_{f} H) C_{2} H_{2} = 225 {kj mol}^{- 1}$
For enthalpy of combustion the corresponding chemical reaction is,
$C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 C O_{2} + H_{2} O$
$Δ H^{\circ} = (2 \times (- 390) - 280 - 225) kj mo l^{- 1} = - 1285 {kj mol}^{- 1}$
$∴ ∣ ∣ ∣ \frac{Δ H^{\circ}}{257} ∣ ∣ ∣ = ∣ ∣ ∣ - \frac{1285}{257} ∣ ∣ ∣ = 5 {kJmol}^{- 1}$

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Similar questions

Q. Consider the following reactions:

(i) H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l), Δ H = - X_{1} {kJ mol}^{- 1}

(i i) H_{2} (g) + \frac{1}{2} O_{2} (a q) \to H_{2} O (l), Δ H = - X_{2} {kJ mol}^{- 1}

(i i i) C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l), Δ H = - X_{3} {kJ mol}^{- 1}

(i v) C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l), Δ H = X_{4} {kJ mol}^{- 1}

Enthalpy of formation of

H_{2} O (l)

is:

Q. Enthalpy of chemical adsorption,

Δ H

k J m o l^{- 1}

is in the range of:

Q. Given,

H_{2} S_{(g)} \to H S (g) + H (g); △ H_{r x n} = 379.6 k J m o l^{- 1}

H_{2} (g) + S (s) \to H_{2} S (g);

△ H_{f}^{\circ} H_{2} S = - 21.1

k J m o l^{- 1}

H_{2 (g)} \to 2 H_{(g)}

;

Δ H = 438 k J m o l^{- 1}

Δ H_{s u b l i m a t i o n}

of S =

280 k J m o l^{- 1}

Thus,

△ H_{f}^{\circ} H S

Q. Consider the following reactions :

H^{+} (a q) + O H^{-} (a q) = H_{2} O (l); Δ H = - X_{1} k J . m o l^{- 1}

II)

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H = - X_{2} k J . m o l^{- 1}

III)

C O_{2} (g) + H_{2} (g) = C O (g) + H_{2} O (l); Δ H = + X_{3} k J . m o l^{- 1}

IV)

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O (g) + H_{2} O (l); Δ H = + X_{4} k J . m o l^{- 1}

Enthalpy of formation of

H_{2} O (l)

is:

Q. If

Δ U

of combustion of

C H_{4} (g)

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

at certain temperature is

- 393 k J m o l^{- 1}

. The value of

Δ H

is:

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Given, C2H2(g)→C2H4(g):ΔH∘=−175 kJ mol−1 ΔH∘f(C2H4,g)=50kJ mol−1;ΔH∘f(H2O,l)=−280 kJ mol−1;ΔH∘f(CO2g)=−390 kJ mol−1 If ΔH∘ is enthalpy of combustion (in kJ mole−1 of C2H2(g), then calculate the value of ∣∣∣ΔH∘257∣∣∣