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murzikaleks [220]
2 years ago
11

In a famous experiment, Wöhler heated ‘inorganic’ ammonium cyanate in the absence of air. The only product of the reaction was ‘

organic’ urea, CO(NH2)2. No other products were formed in the reaction. What is the formula of the cyanate ion present in ammonium cyanate? A CNO– B CNO2– C CO– D NO– why A?
Chemistry
1 answer:
Soloha48 [4]2 years ago
5 0
<span>CO(NH2)2 the only product consists of 1 Carbon 1 Oxygen , 2 Nitrogen and 4 Hydrogen We are heating Ammonium Cyanate, Ammonium Ion NH4+ NH4+ consists of 1 Nitrogen 4 Hydrogen. Compare the reactant decomposed and product formed.

There should be 2 Nitrogen so Cyanate must contain one Nitrogen atom, one Carbon and one oxygen.</span>
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176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

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a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

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a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

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