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Zielflug [23.3K]
3 years ago
9

The reaction described by the equation O 3 ( g ) + NO ( g ) ⟶ O 2 ( g ) + NO 2 ( g ) has, at 310 K, the rate law rate of reactio

n = k [ O 3 ] [ NO ] k = 3.0 × 10 6 M − 1 ⋅ s − 1 Given that [ O 3 ] = 7.0 × 10 − 4 M and [ NO ] = 5.0 × 10 − 5 M at t = 0 , calculate the rate of the reaction at t = 0 .
Chemistry
1 answer:
UNO [17]3 years ago
6 0

Answer:

Explanation:

Equation of the reaction:

O3(g) + NO(g) ⟶ O2(g) + NO2(g)

From the Rate Law:

R = k[O3][NO]

The order of NO is 1 and that of O3 is also 1, therefore order of reaction = 1 + 1

= 2

Given :

Concentration of O3, [O3] = 7.0×10^-4M

Concentration of NO, [NO] = 5.0×10^-5M

Rate constant, k = 3.0 × 10^6M^-1s^-1

R = 3.0×10^6 × 7.0×10^-4 × 5.0×10^-5

= 0.105 M/s

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Answer : The correct option is, (A) AlCl_3

Explanation :

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The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

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The dissociation of MgSO_4 will be,

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According to stoichiometry :

1 mole of Cl_2 produces = 1 mole of MgCl_2

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