For a reaction at `800^(@)C`
`2NO + 2H_(2) rarr N_(2) + 2H_(2)O`, the following data were obtained:
`|{:(,[NO] xx 10^(-4) (M),[H_(2)] xx 10^(-3)(M),-(1)/(2) (d[NO])/(dt) xx 10^(-4) (mol L^(-1) min^(-1))),(i.,1.5,4.0," " 4.4),(ii.,1.5,2.0," "2.2),(iii.,0.5,2.0," "0.24):}|`
What is the order of this reaction with respect to `NO` and `H_(2)` ? Also calculate the rate constant.

For a reaction 2NO(g) + 2H_(2)(g) rarr N_(2)(g)+2H_(2)O(g) the following data were obtained: |{:(,[NO] (mol L^(-1)),[H_(2)] (mol L^(-1)),"Rate" (mol L^(-1) s^(-1))),(1.,5 xx 10^(-3),2.5 xx 10^(-3),3 xx 10^(-5)),(2.,15 xx 10^(-3),2.5 xx 10^(-3),9xx10^(-5)),(3.,15 xx 10^(-3),10 xx 10^(-3),3.6 xx 10^(-4)):}| (a) Calculating the order of reactions. (b) Find the rate constant. (c ) Find the initial rate if [NO] = [H_(2)] = 8.0 xx 10^(-3) M

At 800^(@)C the rate of reaction 2NO + H_(2) rarr N_(2) + H_(2)O Changes with the concentration of NO and H_(2) are [NO] in M " "[H_(2)]" in M "" "-(1)/(2) (d[NO])/(dt)" in M sec"^(-1) (i) 1.5xx10^(-4)" "4xx10^(-3)" "4.4xx10^(-4) (ii) 1.5xx10^(-4)" "2xx10^(-3)" "2.2xx10^(-4) (iii) 3.0xx10^(-4)" "2xx10^(-3)" "8.8xx10^(-4) (a) What is the order of this reaction ? (b) What is the rate equation for the reaction ? (c) What is the rate when [H_(2)]=1.5xx10^(-3)" M and "[NO]=1.1xx10^(-3) M ?

A+ 2B rarr C+D." If "-(d[A])/(dt)=5xx10^(-4)" mol L"^(-1) s^(-1)," then "-(d[B])/(dt) is :

For the decompoistion of HI at 1000 K(2HI rarr H_(2)+I_(2)) , following data were obtained: |{:([HI] (M),"Rate of decomposition of HI" (mol L^(-1) s^(-1))),(0.1,2.75 xx 10^(-8)),(0.2,11 xx 10^(-8)),(0.3,24.75 xx 10^(-8)):}| The order of reaction is

Calculate the pH of the following solutions: a 10^(-2) M HCI b 10^(-3) M H_(2)SO_(4) c 0.2 xx 10^(-2) M NaOH d 0.3 xx 10^(-3) M Ca (OH)_(2)

For the reaction 2N_(2)O_(5) rarr 4NO_(2)+O_(2) rate of reaction and rate constant are 1.02 xx 10^(-4) and 3.4 xx 10^(-5) sec^(-1) respectively. The concentration of N_(2)O_(5) at that time will be

For the reaction N_(2)+3H_(2) rarr 2NH_(3) , if (d[NH_(3)])/(dt).=4xx10^(-4) mol L^(-1)s^(-1) , the value of (-d[H_(2)])/(dt) would be