How many kilojoules is 1,500,000 calories

bagirrra123 [75]

$176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

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

$\text{NH}_4 \text{Cl} \; (s)$ : $-2 \; a = 1$
$\text{NH}_3\; (g)$ : $-2 \; b = -1$
$\text{HCl} \; (g)$ : $2 \; c = -1$

Thus

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

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

Verify this conclusion against a fourth species involved- $\text{N}_2 \; (g)$ for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

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

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

$\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}$

3 0

3 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

2 years ago

HELP !!!!!!! ASAP !!!!!!!!!!!!!!

Dmitry [639]

The color changes, heat change, smell change, are a few

4 0

3 years ago

Read 2 more answers

Convert 1 bromopropane to bromoethane.

Dovator [93]

Answer:

The Dehydrohaogenation of 1-bromo propane with alcoholic KOH gives propene which on again hydrohalogenation with HBr gives 2-bromo propane due to Markonikove's rule for addition.

Explanation:

6 0

2 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,