Lead Melting <br><br> a<br> Chemical Change<br> b<br> Physical Change

The decomposition reaction of N2O5 in carbon tetrachloride is 2N2O5−→−4NO2+O2. The rate law is first order in N2O5. At 64 °C the

Mademuasel [1]

Answer:

(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]

(b) rate = 1.16 x 10⁻⁴ M/s

(c) rate = 2.32 x 10⁻⁴ M/s

(d) rate = 5.80 x 10⁻⁵ M/s

Explanation:

We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so

(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]

(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s

(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s

(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to

1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s

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Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,