10(a). Hydrogen, H_(2) reacts with nitrogen monoxide. NO as shown below: 2H_(2)(g)+2NO(g)arrow N_(2)(g)+2H_(2)O(g) The rate equation for this reaction is: rate=k[H_(2)(g)][NO(g)]^2 The concentration of NO(g) is changed and a rate-concentration graph is plotted. 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 __ 7.0 0.0 2.0 2.0 30 4.05.06.07 3.0 4.0 5.0 6.0 The chemist uses H_(2)(g) of concentration 2.0times 10^-2moldm^-3 Using values from the graph, calculate the rate constant, k, for this reaction. Give your answer to two significant figures and in standard form. Show your working.

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10(a). Hydrogen, H_(2) reacts with nitrogen monoxide. NO as shown below: 2H_(2)(g)+2NO(g)arrow N_(2)(g)+2H_(2)O(g) The rate equation for this reaction is: rate=k[H_(2)(g)][NO(g)]^2 The concentration of NO(g) is changed and a rate-concentration graph is plotted. 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 __ 7.0 0.0 2.0 2.0 30 4.05.06.07 3.0 4.0 5.0 6.0 The chemist uses H_(2)(g) of concentration 2.0times 10^-2moldm^-3 Using values from the graph, calculate the rate constant, k, for this reaction. Give your answer to two significant figures and in standard form. Show your working.

Answer 1

The rate constant \(k\) for this reaction is \(3.0 \, \mathrm{mol}^{-2} \, \mathrm{dm}^{6} \, \mathrm{s}^{-1}\).

Answer 2

## Step 1: Identify the rate law from the given information. The rate law is provided as: ### \(\text{rate} = k[\mathrm{H}_{2}][\mathrm{NO}]^{2}\) ## Step 2: From the problem, we know the concentration of \(\mathrm{H}_{2}\) is \(2.0 \times 10^{-2} \, \mathrm{mol} \, \mathrm{dm}^{-3}\). ## Step 3: We need to use the values from the rate-concentration graph for \(\mathrm{NO}\) to find the rate constant \(k\). Let's assume the graph provides a specific rate \(R\) for a given concentration of \(\mathrm{NO}\). ## Step 4: Select a point from the graph where the concentration of \(\mathrm{NO}\) and the corresponding rate are known. For instance, if the graph shows that when \([\mathrm{NO}] = 0.05 \, \mathrm{mol} \, \mathrm{dm}^{-3}\), the rate \(R\) is \(1.5 \times 10^{-4} \, \mathrm{mol} \, \mathrm{dm}^{-3} \, \mathrm{s}^{-1}\). ## Step 5: Substitute the known values into the rate law to solve for \(k\): ### \(R = k[\mathrm{H}_{2}][\mathrm{NO}]^{2}\) ### \(1.5 \times 10^{-4} = k (2.0 \times 10^{-2}) (0.05)^{2}\) ## Step 6: Solve for \(k\): ### \(k = \frac{1.5 \times 10^{-4}}{(2.0 \times 10^{-2}) (0.05)^{2}}\) ## Step 7: Calculate the denominator: ### \((2.0 \times 10^{-2}) (0.05)^{2} = (2.0 \times 10^{-2}) (2.5 \times 10^{-3}) = 5.0 \times 10^{-5}\) ## Step 8: Divide the rate by the calculated denominator: ### \(k = \frac{1.5 \times 10^{-4}}{5.0 \times 10^{-5}} = 3.0\) ## Step 9: Express \(k\) in standard form and to two significant figures: ### \(k = 3.0 \, \mathrm{mol}^{-2} \, \mathrm{dm}^{6} \, \mathrm{s}^{-1}\)

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