3C2H2g-C6H6lgiven that GmoC2H2-2-4-105JGmoC6H6- -1-4-105JFind the value of log10k at 25-C

ΔG^o=(G_M^o)_C_6H_6 3(G_M^o)_C_2H_2 = 1.4×10^5 3× 2.4×10^5 = 8.6×10^5Joule 2.303RT log_10k= 8.6×10^5 2.303× 8.314× 298log_10k= 8.6×10^5 log_10k=150.72 ...

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Byju's Answer

Standard XII

Chemistry

Arrhenius Acid

3C2H2g⇌C6H6lg...

Question

$3 C_{2} H_{2} (g) ⇌ C_{6} H_{6} (l)$
given that

$G_{m}^{o} (C_{2} H_{2}) = 2.4 \times 10^{5} J$

$G_{m}^{o} (C_{6} H_{6}) = - 1.4 \times 10^{5} J$

Find the value of ${log}_{10} k at 25^{\circ} C$

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Solution

$Δ G^{o} = {(G_{M}^{o})}_{C_{6} H_{6}} - 3 {(G_{M}^{o})}_{C_{2} H_{2}}$
$= - 1.4 \times 10^{5} - 3 \times 2.4 \times 10^{5}$

$= - 8.6 \times 10^{5} J o u l e$

$- 2.303 R T {log}_{10} k = - 8.6 \times 10^{5}$

$- 2.303 \times 8.314 \times 298 {log}_{10} k = - 8.6 \times 10^{5}$

${log}_{10} k = 150.72$

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Similar questions

C_{6} H_{6} + {H N O}_{3} + H_{2} {S O}_{4} 50^{0} - 60^{0} - ---- \to A S n + H C l - ---- \to B {C H}_{3} C O C l - ----- \to C

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Q. Give the empirical formula of:

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(C_{6} H_{6})

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(C_{6} H_{12} O_{6})

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Q. The enthalpy of formation of

C_{2} H_{2 (g)}

and

C_{6} H_{6 (g)}

298 K

and

230

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respectively. The enthalpy change,

Δ H

for the reaction:

3 C_{2} H_{2 (g)} ⇌ C_{6} H_{6 (g)}

298 K

is:

Q. The reaction

C_{6} H_{6} \to C_{6} H_{5} {C H}_{3}

is called:

C_{6} H_{6} + A + A l {C l}_{3} ⟶ C_{6} H_{5} C O {N H}_{2}

;

Here A is:

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3C2H2(g)⇌C6H6(l) given that Gom(C2H2)=2.4×105J Gom(C6H6)=−1.4×105J Find the value of log10k at 25∘C

ΔGo=(GoM)C6H6−3(GoM)C2H2 =−1.4×105−3×2.4×105 =−8.6×105Joule −2.303RTlog10k=−8.6×105 −2.303×8.314×298log10k=−8.6×105 log10k=150.72

$3 C_{2} H_{2} (g) ⇌ C_{6} H_{6} (l)$
given that

$G_{m}^{o} (C_{2} H_{2}) = 2.4 \times 10^{5} J$

$G_{m}^{o} (C_{6} H_{6}) = - 1.4 \times 10^{5} J$

Find the value of ${log}_{10} k at 25^{\circ} C$